Maxima & Minima

1
MAXIMA & MINIMA
A SHORT NOTES
Arun Umrao
https://sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING

2 Maxima & Minima
Contents
1 Extrema 3
1.1 Maxima & Minima . . . . . . . . . . . . . . . . . . . . . . . . 3
1.1.1 Maximum & Minimum Function . . . . . . . . . . . . . 3
1.2 Some Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2.1 Maxima & Minima Points . . . . . . . . . . . . . . . . 5
1.2.2 Critical Points . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.3 Stationary Points . . . . . . . . . . . . . . . . . . . . . 6
1.2.4 Point of Inflection . . . . . . . . . . . . . . . . . . . . . 6
1.2.5 Extreme Value Theorem . . . . . . . . . . . . . . . . . 8
1.2.6 Global Minimum & Global Maximum . . . . . . . . . . 8
1.2.7 Local Minimum & Local Maximum . . . . . . . . . . . 9
1.2.8 Saddle Point . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3 Extremum Test . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.4 Method of Finding Extremum . . . . . . . . . . . . . . . . . . 11
1.4.1 First Derivative Test . . . . . . . . . . . . . . . . . . . 11
1.4.2 Second Derivative Test . . . . . . . . . . . . . . . . . . 12
1.4.3 Why Second Derivative Required . . . . . . . . . . . . 13
1.5 Extremum in Two-Dimensional Function . . . . . . . . . . . . 15
1.6 Applied Problems . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.6.1 Hypothetical Extrema . . . . . . . . . . . . . . . . . . 29

1.1. MAXIMA & MINIMA 3
1Maxima & Minima
1.1 Maxima & Minima
Maxima and Minima are points where function reaches its maximum and
minimum value respectively. These points are also called extrema. Extrema
are of two types: (i) Global extrema (Absolute extrema) and (ii) Local ex-
trema (Relative extrema). Global extrema are those points whose value is
maximum or minimum on the entire range of the function. Local extrema
are the largest or smallest values of the function in the immediate vicinity.
At extrema, the slope of a function in graph is necessarily zero. Extrema are
also called stationary points or turning points.
1.1.1 Maximum & Minimum Function
Certain functions have maximum and minimum values at their certain points.
At other points function value oscillates between these two extreme values.
Illustrated Example Assume sine function y = a sin(x). The domain of
argument of sine varies from 0 to 2nπ. Range of sine function lies between
-1 to +1 irrespective of value of argument.
For Maximum y y shall be maximum if sin(x) is maximum, i.e. sin(x) = 1
or x = π/2. Now the y is y = a.
For Minimum y y shall be minimum if sin(x) is minimum, i.e. sin(x) =
−1 or x = −π/2. Now the y is y = −a.
1 2 3 4 5 6
x
y
a
−a

4 Maxima & Minima
Solved Problem 1.1 Find the maximum and minimum value of y =
a sin2
(x) + b cos2
(x) and a > b.
Solution The given trigonometric function is y = a sin2
(x)+b cos2
(x) and
it is periodic function. General value of arguments of sine and cosine varies
from 0 to 2nπ. Range of sine and cosine lies between -1 to +1. Hence y shall
continuously changes between its maximum and minimum values. Changing
the function for one trigonometric form
y = a sin2
(x) + b 1 − sin2
(x)

Simplifying it for sine
y = b + (a − b) sin2
(x)
For Maximum y y shall be maximum if sin2
(x) is maximum, i.e. sin2
(x) =
1 or sin(x) = ±1. Here taking only +1, x = π/2. Now the y is
y = b + (a − b) = a
For Minimum y y shall be minimum if sin2
(x) is minimum, i.e. sin2
(x) =
−1 or sin(x) = ±
√
−1. It is imaginary value for sine, hence not acceptable.
So for only acceptable values sin2
(x) = 0, i.e. x = 0. Now the y is
y = b + (a − b) × 0 = b
1 2 3
x
y
a
b
x
y
a
b
1.2 Some Definitions
Following definitions are useful in finding the maximum or minimum value
of a function.

1.2. SOME DEFINITIONS 5
1.2.1 Maxima Minima Points
Maxima and minima points on combining known as extrema points. At
extrema points curves changes it direction. There may one or more than one
extrema points for a given function.
x
y
x = ay2
(0, 0)
b
(a)
x
y
x = a − y2
(0, 2)
b
(b)
Figure 1.1: Function plot.
From figure 1.1(1) point (0, 0) is the point at deepest position. This
point is known as the minima point in the given range of the function f(x).
Similarly from figure 1.1(2), point (0, 2) is the point at maximum height. It
is known as maxima point for given range of the function f(x).
1.2.2 Critical Points
A single variable function f(x) has critical points at all points x0 where
f
′
(x0) = 0
or f(x) is not differentiable at the given point.
x
f(x)
f(x) = 2x2
(0, 0)
b
f
′
(x) = 0
(a)
x
f(x)
f(x) = 4 − x2
(0, 2)
b
f
′
(x) = 0
(b)
Figure 1.2:

6 Maxima Minima
In figure 1.2(1) the tangent at lowest depth point is parallel to the positive
x axis. Similarly in figure 1.2(2) tangent at maximum height of the curve is
parallel to the positive x axis. The points at which tangent is parallel to the x
axis are known as critical points or stationary points. At critical points slope
of tangent f
′
(x) is zero. Multi-variable function f(x, y) has critical points
where either the gradient df(x, y) is zero or partial derivative of function is
not defined at that point, ie ∂f(x, y)/∂x or ∂f(x, y)/∂y is undefined.
1.2.3 Stationary Points
Points at which the derivative of a function f(x) vanishes. For example, at
point x0 if first derivative of function is zero, i.e.
f
′
(x0) = 0
Then point x0 is known as stationary point. This stationary point may be
minimum, maximum or inflection point. Value of function at stationary point
is called stationary value.
Solved Problem 1.2 Find the stationary points of function f(x) = 2x − x2
.
Solution The stationary points of the function f(x) is given by f
′
(x) = 0,
i.e.
f
′
(x) = 2 − 2x = 0 ⇒ x = 1
The stationary point of the given function is x = 1.
1.2.4 Point of Inflection
It is a point on a curve at which the curvature or concavity changes sign from
plus to minus or vice-versa. At reflection point, curve changes from being
concave (concave downward) to convex (concave upward), or vice versa. If
f(x) is k times continuously differentiable in a certain neighbourhood of a
point x with k odd and k ≥ 3, while f(n)
(x0) = 0 for n = 2, . . ., k1 and
f(n)
(x0) 6= 0 then f(x) has a point of inflection at x = x0.

1
−1
1
−1
x
f(x)
b
f(x) = x3
f
′
(x) = 3x2
f
′′
(x) = 6x
Above graph is drawn for f(x) = x3
. The point x at which slope of curve
is parallel to horizontal axis, is found by using
f
′
(x) = 0
It gives x = 0. Now again derivating f
′
(x), we get
f
′
(x) = 6x
f
′
(x) is zero at x = 0, hence x = 0 is point of inflection of given function
f(x). In the following figure, function g(x) = x3
− 3x2
+ 2x is plotted. Its
derivative is g
′
(x) = 3x2
− 6x + 2.
1
−1
1 2
−1
x
g(x)
b
g(x)
g
′
(x)
g
′′
(x)
The point of inflection shall be where, there g
′′
(x) = 0. So,
g
′
(x) = 0 = 6x − 6
It gives, x = 1. This is inflection point for the given function g(x). Note that,
inflection point may or may not be a root point, i.e. zeros of the function.
For example, take
y = x3
− 3x2
− 144x + 432

8 Maxima Minima
Its roots are x = −12, x = 3 and x = 12. The point of inflection will be
found by taking y
′′
= 0. So,
6x − 6 = 0 ⇒ x = 1
which is not root of the given function y.
1.2.5 Extreme Value Theorem
If a function h(x) is continuous on a closed interval [p, q], then h(x) has both
maximum and minimum on closed interval [p, q]. If h(x) has an extremum
on an open interval (p, q) then extremum occurs at a critical point.
1.2.6 Global Minimum Global Maximum
The overall largest value of a set, function, etc., over its entire range is called
global maximum. While smallest value of a set, function, etc., over its entire
range is called global minimum. It is impossible to construct an algorithm
that will find a global maximum or global minimum for an arbitrary function.
We can check whether function has global extrema or not by just putting
x → ±∞. For example,
y = x2
− 2x + 4
global maxima as
lim
x→−∞
y = (−∞)2
− 2 × −∞ + 4 = ∞
and
lim
x→+∞
y = ∞(∞ − 2) + 4 = ∞
Thus function is not converging when x → ±∞. Using these tests, we can
expect local extrema.
1. When limx→−∞ y = limx→+∞ y = ±∞, i.e. both have same values
with then function may have odd numbers of extrema. This happens when
highest degree of function is even.

x
y
x → −∞ x → ∞
x
y
x → −∞ x → ∞
2. When limx→−∞ y = −∞ and limx→+∞ y = +∞ or vice-versa then
function may have even numbers of extrema. This happens when highest
degree of function is odd.
x
y
x → −∞ x → ∞
x
y
x → −∞ x → ∞
1.2.7 Local Minimum Local Maximum
It is also known as relative minimum and relative maximum. As-
sume a function f(x), whose successive derivatives at point x0 are zero. i.e.
f
′
(x0) = 0, f
′′
(x0) = 0, . . . . . . , f(n)
(x0) = 0
But
f(n+1)
(x0) 6= 0
Now, f(x) has a local maximum at x0 if n is odd and f(n+1)
(x0) 0. Sim-
ilarly, f(x) is local minima at x0 if n is odd and f(n+1)
(x0) 0. The (n)th
derivative of function is zero at point x0 but (n)th
derivative of function is
not zero at point x0 hence it gives the maximum and minimum point. To
find the saddle point we have to find the maximum derivatives of function
that gives zero value at point x0 and also check is n odd or even. In this
case, highest derivative that gives zero value at x0 has derivative power n.
There is a saddle point at x0 if n is even.
Bounding Values If there are extrema points within closed range [a, b],
then the extreme values (positive or negative) of the function are also known

10 Maxima Minima
as bounding values of the function. For example, suppose function
f(x) = x − x2
which is continuous and differentiable at every point defined within (0, 1).
The function has a extrema at
f
′
(x) = 1 − 2x = 0 ⇒ x = 0.5
Now the function value at x = 0.5 is
f(0.5) = 0.25
Again second derivative
f
′′
(0.5) = −2
is negative value. Hence the extrema is also local maxima. Now the roots
of function are
x − x2
= 0 ⇒ x = 0, 1
Function value at these two points are
f(0) = 0 f(1) = 0
Hence the function value ranges between (0, 0.25) ie in other words the
function is bounded between 0 to 0.25 (upper bound by 0.25 and lower
bound by 0).
1.2.8 Saddle Point
A point on a function f(x) which is a stationary point but not an extremum.
To clarify the point assume an function f(x) = 3x3
. Saddle point is mostly
visible in one-dimensional functions with one term. First derivative of the
function is f
′
(x) = 9x2
. Second derivative of the function is f
′′
(x) = 18x.
And third derivative is f
′′′
(x) = 18. The function has a saddle point at
x0 = 0 by extremum test as1
f
′′
(x0) = 0; f
′′′
(x0) = 18
1
3rd
derivative has non zero value at point x0 while 2nd
derivative has zero value at
x0. 2 power in second derivative is an even number hence there is a saddle point at x0.

1.3. EXTREMUM TEST 11
Where, third derivative is not zero at point x0.
1.3 Extremum Test
To find the extrema of a function f(x) at all points x0, first derivative of func-
tion f(x) is obtained. As the slope must be zero at stationary points hence
the turning points can be calculated by equating first derivative to zero, ie
f
′
(x) = 0. This gives all the values of points x0. To classify a stationary
point extremum test or second derivative test is carried out. If second deriva-
tive of the function f(x) at extremum points is positive (f
′′
(x0) 0) then
gradient is increasing and stationary point is minimum. If second derivative
of the function f(x) is negative (f
′′
(x0) 0) then gradient is decreasing and
stationary point is maximum. If second derivative is zero at stationary point
(f
′′
(x0) = 0) then it is a point of inflection or it could be an extremum. To
find the extrema we differentiate the first derivative until we would not find
the non zero derivative at that stationary point.
1.4 Method of Finding Extremum
There are two methods of finding extremum. (i) First derivative test and (ii)
second derivative test.
1.4.1 First Derivative Test
Suppose a function f(x) that is continuous at a stationary point x0. First
derivative of the function f(x) represents the slope of tangent on the function
at the point ‘x’. For maxima or minima at any point on the function graph,
following rules are observed.
1. If f
′
(xl) 0 on an open interval extending left from x0 and f
′
(xr) 0
on an open interval extending right from x0, then f(x) has a local
maximum at x0. It is also possibility that there is an global maximum.

12 Maxima Minima
x
y
b
b
b
xl x0 xr
f
′
(xl) 0
f
′
(x0) = 0
f
′
(xr) 0
+θ
−θ
2. If f
′
(xl) 0 on an open interval extending left from x0 and f
′
(xr) 0
on an open interval extending right from x0, then f(x) has a local
minimum at x0. It is also possibility that there is an global minimum.
x
y
b
b
b
xl x0 xr
f
′
(xl) 0
f
′
(x0) = 0
f
′
(xr) 0
−θ
+θ
3. If f
′
(x) has the same sign on an open interval extending left from x0
and on an open interval extending right from x0, then f(x) has a point
of inflection at x0.
1.4.2 Second Derivative Test
Suppose a function f(x) is continuous and twice differentiable at stationary
point x0. There are two cases of the f
′′
(x) at point x0.
1. If f
′′
(x0) 0, then slope of tangent on function f
′
(x0) is upward and
the function f(x0) has a relative minima at stationary point x0.
2. If f
′′
(x0) 0, then slope of tangent on function f
′
(x0) is downward
and the function f(x0) has a relative maxima at stationary point x0.

1.4. METHOD OF FINDING EXTREMUM 13
x
y
b
b
b
xl x0 xr
f
′
(xl)
f
′
(x0)
f
′
(xr)
(a)
x
y
b
b
b
xl x0 xr
f
′′
(xl) f
′′
(x0)
f
′′
(xr)
(b)
−θ
(c)
Figure 1.3: Figure (a) graphed presentation of first derivative of sin(x), (b)
graphed presentation of second derivative of sin(x) and (c) slope of second
derivative of sin(x) at point x0.
1.4.3 Why Second Derivative Required
Take a function f(x) whose curve is shown in the following figure. It has
four inversion points at x = −2, x = −1, x = 0, x = 1 and x = 2. First
derivative of function f(x) is plotted too. f
′
(x) is plot of numeric values of
the tangents on functions f(x). It means, f
′
(x) should have zero numeric
value where function f(x) has function inversion. This is why, to get the
inversion points of function f(x), first derivative of function should be equal
to zero. It gives all inversion points but it does not tell about the point that
whether it is maxima or it is minima. The first derivative of function, i.e.
f
′
(x) clearly tells about the nature of function. When f
′
(x) 0, function is
increasing and when f
′
(x) 0 function is decreasing.
x
f(x)
f
f
′
x
=
1.58
x
=
0
x
=
−1.58
(a)
x
f(x)
f
f
′
f
′′
x
=
1.58
x
=
0
x
=
−1.58
(b)
When function changes from decreasing to increasing, the numeric value
of f
′
(x) changes from negative value to positive value. It means, f
′
(x) has
zero crossing at point of inversion and function plot has slop too. If this slope
is positive then f
′′
(x) should be positive value. Similarly, when function
changes from increasing to decreasing, the numeric value of f
′
(x) changes

14 Maxima Minima
from positive value to negative value. It means, f
′
(x) has zero crossing at
point of inversion and function plot has slop too. If this slope is negative then
f
′′
(x) should be negative value. Hence, where f
′′
(x) is negative, then at that
point function has maxima and where f
′′
(x) is positive, then at that point
function has minima. This is the reason, that second derivative is required
to verify the maxima and minima point.
Solved Problem 1.3 A company produces auto parts and earns income from
sale of these parts. The earning from sale of auto parts is based on the labour
(l) employed and capital (c) invested by the company. Earning function is
I = 2l + 4c − 3l2
− 2c2
+ 9lc
Find the quantity of labour employed and amount of capital invested by the
company to earn the maximum income.
Solution The variation in the income of the company when there is
variation in labour employed or capital invested is maximum or minimum
when dI/dl = 0 and dI/dc = 0. So,
dI
dl
=
d
dl
2l + 4c − 3l2
− 2c2
+ 9lc

= 2 − 6l + 9c = 0
And
dI
dc
=
d
dc
2l + 4c − 3l2
− 2c2
+ 9lc

= 4 − 4c + 9l = 0
1
0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
l c
I
l c
Two relations of labour and capital are
6l − 9c = 2; 9l − 4c = 4
On solving these equations, we have l = 68/171 and c = 2/19. Again, we
check whether income is maximum at these values of labour employed and

1.5. EXTREMUM IN TWO-DIMENSIONAL FUNCTION 15
capital invested. So, we shall find d2
I/dl2
and d2
I/dc2
. So,
d2
I
dl2
= −6;
d2
I
dc2
= −4
As both are negative, hence, income shall be maximum at this labour em-
ployed and capital invested.
1.5 Extremum in Two-Dimensional Function
If f(x, y) is a two-dimensional function that has a relative extremum at a
point (x0, y0) and has continuous partial derivatives at this point, then
∂f(x0, y0)
∂x
= 0
and
∂f(x0, y0)
∂y
= 0
The second partial derivatives test classifies the point as a local maximum
or relative minimum. The second derivatives of the function is D then
D = fxxfyy − f2
xy
Now
1. If D 0 and fxx(x0, y0) 0, the point is a relative minimum.
2. If D 0 and fxx(x0, y0) 0, the point is a relative maximum.
3. If D 0, the point is saddle point.
4. If D = 0, higher order tests must be used.
Note that, if we have not given the points x0 and y0 then we can find these
points where extremum may exist by solving the equations obtained by using
relation ∂f/∂x = 0 and ∂f/∂y = 0.

16 Maxima Minima
1.6 Applied Problems
Solved Problem 1.4 Find the maximum value of y = x3
− 9x.
Solution The given function y is zero when x = 0. To get the critical
points where tangent on the function is parallel to abscissa, we derivative the
function with respect to variable ‘x’. Now
dy
dx
= 3x2
− 9
0 = 3x2
− 9
On solving, the critical points are x = ±
√
3.
x
f(x) f
(
x
)
=
x
3
−
9
x
b
f
′
(x) = 0
b
f
′
(x) = 0
(a)
x
f(x)
f
(
x
)
=
x
3
−
9
x
b
f(−
√
3)
b
(−
√
3, 0)
b
f(
√
3)
b
(
√
3, 0)
(b)
Figure 1.4:
From figure 1.4(b), function is continuously decreasing when x ≤ −
√
3
and continuously increasing when x ≥
√
3. In the domain of −
√
3 ≤ x ≤
√
3
function is decreasing. To verify that these points are at minima or maxima
location, second derivative value of the function y is obtained at these critical
points.
d2
y
dx2
= 6x
At x = −
√
3, second derivative is negative while at x =
√
3, second derivative
is positive. Now the maxima exists at x = −
√
3 and maximum function value
is
y−
√
3 = −3
√
3 + 9
√
3 = 6
√
3
This is the local maximum value of the function. But from the figure 1.4(b)
it is clear that the minimum value of the function is −∞ and maximum value
of the function is +∞ globally.

1.6. APPLIED PROBLEMS 17
Solved Problem 1.5 Find the maximum value of f(x) = 6x − x2
.
Solution The given function is f(x) = 6x − x2
. To find the maximum
value of the function, we must knew the point where function is maximum.
For this, f
′
(x) = 0.
6 − 2x = 0 ⇒ x = 3
There is only one extrema point.
x
f(x)
f
(
x
)
=
6
x
−
x
2
b
f
′
(x) = 0
(a)
x
f(x)
f
(
x
)
=
6
x
−
x
2
b
f(3)
b
(3, 0)
(b)
At x = 3 function shall be either minima or maxima. To check the
extremum of the function, we shall find f
′′
(x)

x=3
. Now,
f
′′
(3) = −2 0
Hence function is maxima at x = 3. Now the function value at x = 3 is
f(3) = 6 × 3 − 32
= 9
This is result. Note that, function has only one critical point where extrema
function is maximum. Hence there is no definite minimum function value.
Solved Problem 1.6 Find the minimum value of f(x) = x2
− 4x + 4.
Solution The given function is f(x) = x2
−4x+ 4. To find the minimum
value of the function, we must knew the point where function is minimum.
For this, f
′
(x) = 0.
2x − 4 = 0 ⇒ x = 2
There is only one extrema point.

18 Maxima Minima
x
f(x)
b
f
′
(x) = 0
(a)
x
f(x)
b
f(2)
b
(2, 0)
(b)
At x = 2 function shall be either minima or maxima. To check the
extremum of the function, we shall find f
′′
(x)

x=2
. Now,
f
′′
(2) = 2 0
Hence function is minima at x = 2. Now the function value at x = 2 is
f(2) = 22
− 4 × 2 + 4 = 0
This is result. Note that, function has only one critical point where extrema
function is minimum. Hence there is no definite maximum function value.
Solved Problem 1.7 Find the minimum value of f(x) = x
4
− x3
5
.
Solution The given function is
f(x) =
x
4
−
x3
5
This function has extremum where f;
(x) = 0. So,
1
4
−
3x2
5
= 0 ⇒ 5 − 12x2
= 0
It gives x = −0.65 and x = +0.65. This function has two extrema. Hence
we shall test whether function minima or function maxima exits.
x
f(x)
b
f
′
(x) = 0
b
f
′
(x) = 0
(a)
x
f(x)
b
f(0.65)
b
(0.65, 0)
b
f(−0.65)
b
(−0.65, 0)
(b)

To test the extrema, we shall find the f
′′
(x) at x = ±0.65. So,
f
′′
(x) = −
6x
5
f
′′
(x) is negative at x = 0.65, hence function maxima is at x = 0.65. f
′′
(x)
is positive at x = −0.65. hence function is minima at x = −0.65. The
minimum function value is
f(−0.65) =
−0.65
4
−
(−0.65)3
5
= 0.107575
and the function maximum value is
f(0.65) =
0.65
4
−
(0.65)3
5
= 0.107575
Hence, function minimum value is −0.10 approximately.
Solved Problem 1.8 Find the maximum value of f(x) = 2
√
x − 2x + 3.
Solution To get the critical points of the function, we should put f
′
(x) =
0. So,
2 ×
1
2
×
1
√
x
− 2 = 0
This give √
x = 0.5 ⇒ x = 0.25
x
f(x)
b
f
′
(x) = 0
f
=
2 √
x
−
2x
+
3
(a)
x
f(x)
b
f(1/4)
b
(1/4, 0)
f
=
2 √
x
−
2x
+
3
(b)
This has only one x = 0.25, where one extrema exists. To check whether
function is maximum or minimum at this point, we shall find f
′′
(0.25). So,
f
′′
(x) = −
1
2
×
1
3
√
x
0; ∀x ∈ R+

20 Maxima Minima
At x = 0.25, f
′′
(0.25) is negative, hence function is maxima at x = 0.25. The
function value at this point is given by
f(0.25) = 2 ×
√
0.25 − 2 × 0.25 + 3 = 3.5
This is desired result.
Solved Problem 1.9 Find the maximum value of f(x) = 6x − x2
.
Solution To get the critical points of the function, we should put f
′
(x) =
0. So,
6 − 2x = 0 ⇒ x = 3
x
f(x)
f(x) = 6x − x2
b
f
′
(x) = 0
(a)
x
f(x)
f(x) = 6x − x2
b
f(3)
b
(3, 0)
(b)
This has only one x = 3, where one extrema exists. To check whether
function is maximum or minimum at this point, we shall find f
′′
(3). So,
f
′′
(x) = −2 0; ∀x
At x = 3, f
′′
(3) is negative, hence function is maxima at x = 3. The function
value at this point is given by
f(3) = 6 × 3 − 32
= 9
Solved Problem 1.10 Find the minimum value of f(x) = cos(3x) in range of
[0, π/2].
Solution The given function is f(x) = cos(3x). The tangents on the
function at its extrema will be there where f
′
(x) = 0. So,
−3 sin(3x) = 0 ⇒ sin(3x) = 0

Note that, function is a periodic function, hence it shall have infinite numbers
of extrema and corresponding critical points. But we are taking here only
local extrema, i.e. maximum and minimum values within definite domain.
x
f(x)
f(x) = cos(3x)
b
b
f
′
(x) = 0
f
′
(x) = 0
(a)
x
f(x)
f(x) = cos(3x)
b
b
f (0)
f (1.046)
b b
(0, 0)
(1.046, 0)
(b)
Now, 3x = nπ or x = nπ/3. We have to find the minimum value in
x ∈ [0, π/2]. For this domain, only accepted critical points2
are found when
n = 0 and n = 1. So, x is 0 and π/3. To check where, function is minimum,
we have to find f
′′
(x) 0. Now
f
′′
(x) = −9 cos(3x)
f
′′
(x) is positive when cos(3x) is negative, i.e. π/2 ≤ 3x ≤ π or π/6 ≤ x ≤
π/3. Note that we are talking about cos(3x) of f
′′
(x) not about the cos(3x)
of f(x). So, only acceptable value for minima from x = 0 and x = π/3 is
x = π/3. Now minimum function value is
f(π/3) = cos

3 ×
π
3

= −1
Solved Problem 1.11 Find the maximum value of y = sin(x) + tan(x) in
range of (−2π, 2π).
Solution The given function is y = sin(x) + tan(x). To get the critical
points, we should put y
′
(x) = 0. Now,
cos(x) + sec2
(x) = 0
On solving it, we have
cos3
(x) = −1
2
Roots are those points where function is zero. Critical points are those points where
tangent on the function is zero.

22 Maxima Minima
This has three roots, in which only one is real. The real root is cos(x) = −1.
cos(x) = cos(2n + 1)π ⇒ x = (2n + 1)π
Here, n is integer. The values of x withing (−2π, 2π) is −π and π only when
n = −1 and n = 0. These are only points where extrema exist.
x
y(x)
y(x) = sin(x) + tan(x)
b b
y
′
(x) = 0 y
′
(x) = 0
(a)
x
y(x)
y(x) = sin(x) + tan(x)
f (−π) f (π)
b b
b b
(−π, 0) (π, 0)
(b)
To check the nature of extrema, we shall get the value of y
′′
(x) at x = −π
and x = π. Hence
y
′′
(x) = 2 sec2
(x) × tan(x) − sin(x)
At x = −π
y
′′
(−π) = 2 sec2
(−π) × tan(−π) − sin(−π) = 0
At x = π
y
′′
(π) = 2 sec2
(π) × tan(π) − sin(π) = 0
At both critical points, y
′′
(x) is zero. Hence function has reflection at points
x = ±π.
Solved Problem 1.12 Find the range of θ for sine function of unit amplitude
so that the bounded rectangular area is maximum.
Solution
θ
y
b
θ
b
f(θ)
b
π − θ
b
f(θ)
θ
y

As per the symmetry of the sine function of unit amplitude, shape of
maximum area will be in rectangular shape as shown in above figure. Area
of this shape is
A = f(θ) × (π − 2θ)
On derivating it with respect to θ and equating it to zero, it gives
tan(θ) + θ −
π
2
= 0
The solution of this relation gives the critical value of θ within the limit of
0 θ π is θ = 0.7104. Second derivative of the area is
A′′
= − [cos θ − (π − 2θ) sin θ − 2 cos θ]
It is negative when θ is substituted. Area will be maximum at the criti-
cal value of θ = 0.7104. Maximum area is bounded between the limits of
0.7104 ≤ θ ≤ 2.4311.
Solved Problem 1.13 Find the minimum and maximum value of f(x) =
sin2
(x) in range of [0, π/2].
Solution The critical points of the function are found by putting f
′
(x) =
0. So,
2 sin(x) × cos(x) = 0 ⇒ sin(2x) = 0
This gives
sin(2x) = sin(nπ) ⇒ x = n
π
2
The values of x lies in [0, π/2] are x = 0 and x = π/2.
x
f(x)
f(x) = sin2
(x)
b
b
f
′
(x) = 0
f
′
(x) = 0
(a)
x
f(x)
f(x) = sin2
(x)
b
b
f π
2

f (0)
b
b
π
2
, 0

(0, 0)
(b)
The extrema of function at these critical points are either maxima or
minima. To check this, we have to find sign of f
′′
(x).
f
′′
(x) = 2 cos(2x)

24 Maxima Minima
At x = 0, f
′′
(0) = 2 0, hence function has minima at x = 0. Similarly at
x = π/2, f
′′
(π/2) = −2 0, hence function has maxima at x = π/2. The
minima and maxima values are respectively
f(0) = [sin(0)]2
= 0
and
f(π/2) = [sin(π/2)]2
= 1
Solved Problem 1.14 Find the minimum and maximum value of f(x) =
sin(x2
) in range of [0, π/2].
Solution To get the extrema of the function, we should put f
′
(x) = 0.
Now, it gives
cos(x2
) × 2x = 0
This gives x = 0 and cos(x2
) = cos((2n + 1)π/2). Or
x2
=
(2n + 1)π
2
This gives,
x = ±
r
(2n + 1)π
2
x
f(x)
f(x) = sin(x2
)
b
b
b
f
′
(x) = 0
f
′
(x) = 0
f
′
(x) = 0
(a)
x
f(x)
f(x) = sin(x2
)
b
b
b
f

−
q
π
2

f (0)
f
q
π
2

b b b

−
q
π
2
, 0

(0, 0)
q
π
2
, 0

(b)
Values of x in the domain [0, π/2] are x = 0 and at n = 0. x = +
p
π/2.
x = −
p
π/2 is not acceptable as it is out of domain. To check where function
is maximum and where function is minimum, we have to check sign of f
′′
(x).
So
f
′′
(x) = cos(x2
) × 2 + 2x × − sin(x2
) × 2x
Or
f
′′
(x) = 2 cos(x2
) − 4x2
sin(x2
)

This value at x = 0 and x =
p
π/2 are
f
′′
(0) = 2 cos(02
) − 4 × 02
× sin(02
) = 2 0
f
′′
(
p
π/2) = 2 cos(π/2) − 4 ×
π
2
× sin(π/2) = −2π 0
Hence function is maximum at x =
p
π/2 and minimum at x = 0. So
maximum and minimum function values are respectively:
f
r
π
2

= sin
π
2

= 1
f(0) = sin(0) = 0
Solved Problem 1.15 Is it possible to find the maximum or minimum values
of the function f(x) = tan(2x) in range of [0, π/2]?
Solution If tangents on the curve are not parallel to horizontal axis in
the given domain [0, π/2], then function is not changing its direction from
increasing to decreasing or vice-versa in the given domain [0, π/2]. If there
is a point within the domain [0, π/2] at which tangent is parallel to x-axis,
then we must put f
′
(x) = 0 and find the critical points. If there are nay
any critical point within the domain [0, π/2], then there is no tangent that
is parallel to horizontal axis and exists within the domain [0, π/2].
x
f(x)
f(x) = tan(2x)
b b
(0, 0)
π
2
, 0

So,
f
′
(x) = sec2
(2x) × 2 = 0
This gives
sec2
(2x) = 0 ⇒ 1 + tan2
(2x) = 0

26 Maxima Minima
Or
tan2
(2x) = −1 ⇒ tan(2x) =
√
−1 = i
Now, there are no points of x within given domain for which tangent satisfy
above condition. Hence tangents on the given function are not parallel to
the x-axis. Hence it is not possible to find the maximum or minimum value
of the given function at given domain.
Solved Problem 1.16 Is function f(x) = tan(2x) is strictly increasing in
range of [0, π/2]?
Solution For any two points a, b ∈ I, if a function f(x) satisfy the
relation f(b) f(a), then it is said to be strictly increasing function for all
b a. If function has f(b) ≥ f(a) for any two values of a and b then we
said that function is nonstrictly increasing. So, for the given function, we
first check whether there is minima or maxima of the function in [0, π/2]. If
there is so, two points either side of the extrema shall give equal values of
function. So,
f
′
(x) = 2 sec2
(2x) = 0
The range of secant is [1, ∞). So, for any value of x, there is never zero value
of secant. So, no point exits in [0, π/2] at which there is extrema on the given
function f(x). Now, for two extreme values, x = 0 and x = π/2, function
values are
f(0) = tan(0) = 0; f(π/2) = tan(2 × π/2) = tan(π) = 0
Here, two values f(0) and f(π/2) are equal, hence function is not strictly
increasing in the given domain x ∈ [0, π/2].
Solved Problem 1.17 Income of a company is represented by I = x3
−cx+3.
Find the maximum income of the company when c = 3.
Solution The income of a company is given by
I = x3
− cx + 3
This income curve shall be maximum or minimum, where I
′
= 0. So,
3x2
− c = 0 ⇒ ±x =
r
c
3

The income shall be maximum at x if I
′′
(x) 0. So,
I
′′
(x) = 6x
This is negative at x = −

, hence income is maximum at x = −

.
Substituting c = 3, we have, x = −1 Now, maximum income of the company
is
Imax = (−1)3
− 3 × −1 + 3 = 5
This is maximum income of the company.
Solved Problem 1.18 Marks of a student is represented by M = 9x2
−3x+5.
Find the maximum marks of the student.
Solution The maximum marks is given by
M = 9x2
− 3x + 5
This mark curve shall be maximum or minimum, where M
′
= 0. So,
18x − 3 = 0 ⇒ x =
1
6
The marks shall be maximum at x if M
′′
(x) 0. So,
M
′′
(x) = 18 0
This is positive at x = 1/6 and it is minima condition. Hence there is no
upper limit of maximum marks. So, this marks curve tells that student has
no definite maximum marks.
Solved Problem 1.19 Sum of two arbitrary numbers is 24. Find the minimum
value of product of these two arbitrary numbers if product is given by p =
xy2
.
Solution The two numbers are x and y. According the question, x+ y =
24. The product function is
p = xy2
= (24 − y)y2
This product curve shall be maximum or minimum, where p
′
= 0. So,
48y − 3y2
= 0 ⇒ 3y(16 − y) = 0

28 Maxima Minima
This gives, y = 0 or y = 16. y = 0 is not acceptable as it make product zero.
Other number is y = 16. We shall check whether product is maximum for
y = 16. For this p
′′
(y) 0. So,
p
′′
(y) = 48 − 6y
This is negative, at y = 16, hence product shall be maximum at y = 16. Now
the other number is x = 24 − 16 = 8. The product is
p = 8 × 162
= 2048
Solved Problem 1.20 Velocity of a moving particle in horizontal plane is
represented by v = x3
− 4x. Here x is an arbitrary parameter. Find the
maximum velocity of the particle.
Solution The velocity function is
v = x3
− 4x
This velocity curve shall be maximum or minimum, where v
′
= 0. So,
3x2
− 4 = 0 ⇒ ±x =
r
4
3
We shall check where velocity is maximum. For this v
′′
(x) 0. So,
v
′′
(x) = 6x
v
′′
(x) is negative at x = −
q
4
3
, hence velocity shall be maximum at this
point. Now, velocity is
v = −
r
4
3
!3
− 4 × −
r
4
3
=
16
3
√
3

1.6.1 Hypothetical Extrema
There are some hypothetical extrema exist in a function. For example, a
time function h = t3
− 6t − 8 has an exremum point at t = −
√
2. But in
timeline, this value never exits as time always started at t = 0. See the
following example:
Solved Problem 1.21 A container is filled by water dripping from a tap. The
height of water in container is a time function and it is given by h = t3
−6t+8.
Find the maximum rate of water filling to the container.
Solution The water filling function is
h = t3
− 6t − 8 = f(t)
This function gives f(−∞) = −∞ and f(∞) = ∞. Hence this function have
global and may have local extrema. Function values at local extrema tell us
whether water filling is maximum or minimum in definite time duration. So,
to find the critical points of extrema, we have f
′
(t) = 0. So,
3t2
− 6 = 0
It gives two critical points, t =
√
2 or t = −
√
2.
t
f(t)
f
(
t
)
=
t
3
−
6
t
−
8
b
b
f
′
(t) = 0
f
′
(t) = 0
(a)
t
f(t)
f
(
t
)
=
t
3
−
6
t
−
8
b
b
f(
√
2)
f(−
√
2)
b
b
√
2, 0

−
√
2, 0

(b)
To check whether function is maxima or minima at a given critical point,
f
′′
(t) 0 or f
′′
(t) 0 respectively. Now
f
′′
(t) = 6t
This is negative at t = −
√
2 and at this point filling function shall be maxi-
mum. So filling is maximum at time t = −
√
2. This time never exists in time
line, hence maximum water filling rate shall never achieved at local extrema.

30 Maxima Minima
Solved Problem 1.22 A moving distance relation for a particle is y = t3
−
6 ∗ t2
+ 11 ∗ t − 6. Find the extremum speed of the particle if possible.
Solution The distance relation is y = t3
−6t2
+ 11t−6. Velocity relation
will be derivative of the distance relation y. So,
v =
d
dt
y = 3t2
− 12t + 11
This velocity curve has extrema points. To find these extrema points, we
will get v
′
(t) = 0. So
v
′
(t) = 6t − 12 = 0
This gives t = 2, There is only one extrema point, hence there will be either
minima or maxima at this point.
x
y(x) v(x)
y(x)
b
v
′
(x) = 0
(a)
x
y(x) v(x)
y(x)
b
v(2)
b
(2, 0)
(b)
To check, there is maxima or minima, get v
′′
(t).
v
′′
(t) = 6 0
This is positive value irrespective of t. So, there shall be minimum velocity
at t = 2. The minimum velocity is
v(2) = 3 × 22
− 12 × 2 + 11 = −1
This is desire result.
Solved Problem 1.23 Find the dimensions of rectangle inscribed in semi-
circle of radius r having maximum area.
Solution Let there is semi-circle of radius r. A rectangle is inscribed in-
side this semi-circle as shown in first part of below figure. Let the dimensions
of the rectangle are 2x and y.

1
1
−1
−2
x
y
b
r
x
y
2
−2
1
−1
x
A
Now area of the rectangle is
A = 2xy
This is two variable relation. We can convert it to one variable relation if x
and y are given in relation with r (constant). From the first part of above
figure
r2
= x2
+ y2
Replacing y in the area relation, we have
A = 2x
√
r2 − x2
The area is maximum, where x gives maxima point on the area curve. To
get critical points, put A
′
(x) = 0.
A
′
(x) = 2x ×
1
2
× (r2
− x2
)−1/2
× −2x + 2
√
r2 − x2 = 0
x2
=
r2
2
⇒ ±x =
r
√
2
⇒ x = ∓
r
√
2
1
1
−1
−2
x
y
b
r
x
y
1
1
−1
−2
x
y
√
2 r
r/
√
2
To check whether area is maximum or minimum at x = ∓ r
√
2
, we have to
find the sign of A
′′
(x) at the given x points.
A
′′
(x) = −
6x
√
r2 − x2
−
2x3
(r2 − x2)
3
2

32 Maxima Minima
Value of A
′′
(x) at x = ∓ r
√
2
are
A
′′

r
√
2

= −
6 ×

r
√
2

r
r2 −

r
√
2
2
−
2 ×

r
√
2
3

r2 −

r
√
2
2
3
2
0
At this point area is maximum. Similarly,
A
′′

−
r
√
2

= −
6 ×

− r
√
2

r
r2 −

− r
√
2
2
−
2 ×

− r
√
2
3

r2 −

− r
√
2
2
3
2
0
At this point area is minimum. Now, the maximum area of the inscribe
rectangle is at x = r
√
2
. Now, the maximum area of inscribed rectangle is
A = 2 ×
r
√
2
×
s
r2 −

r
√
2
2
= r2
The dimensions of the rectangle are 2x =
√
2 r and y =
√
r2 − x2 = r/
√
2.
Solved Problem 1.24 Cost of production of x articles is given by c(x) = 2x3
−
4x2
+4 and revenue received from the selling of same articles is r(x) = x2
−1.
Find the production of articles when profit is maximum.
Solution

x
p(x)
c(x) = 2x3
− 4x2
+ 4
r(x) = x2
− 1
b
b
b
c
′
(x) = 0
c
′
(x) = 0
r
′
(x) = 0
(a)
x
p(x)
c(x) = 2x3
− 4x2
+ 4
r(x) = x2
− 1
b
b
b
c(0)
c(1.33)
r(0)
b b
b
(0, 0)
(1.33, 0)
(b)
Figure 1.5:
If cost of production of ‘x’ items is c(x) and revenue received from selling
of ‘x’ articles is r(x) then profit of selling of ‘x’ items is p(x) = r(x) − c(x).
If r(x), c(x) are differentiable for all x 0 then maximum profit occurs at
p
′
(x) = 0. Now
d p(x)
dx
=
d r(x)
dx
−
d c(x)
dx
At maximum profit
d p(x)
dx
= 0
ie
d r(x)
dx
=
d c(x)
dx
Substituting the values of r(x) and c(x), we have
2x = 6x2
− 8x
On solving it
x = 0 or
=
5
3
≈ 1.666
The second derivatives of r(x) and c(x) at this value of ‘x’ are
d2
r(x)
dx2
= 2 = +ve

34 Maxima Minima
and
d2
c(x)
dx2
= 12x − 8 = +ve
respectively. The profit is maximum when production units are near about
1.666 in factor.
Solved Problem 1.25 A cabinetmaker needs wood to produce 5 furnishings
each day. He orders wood from woodseller periodically. The delivery cost
of wood is Rs. 5000 per supply order. The storage cost of one furnished
cabinet is Rs. 10 per day. Considering the delivery cost and storage charges,
calculate the period after which he should order to minimize his average daily
cost. Also calculate the quantity of wood to be delivered in one delivery.
Solution
x
c(x)
y = 5000
x
y = 25 · x
c(x) = 5000
x
+ 25 · x
x-min
Figure 1.6:
Assume the cabinetmaker should order wood after ‘x’ days. In ‘x’ days
he will produce total furnishings ‘5x’. The average cost of storage S(x) of
furnished cabinet for ‘x’ days is product of furnished cabinets, number of
days to be stored, cost of storage per unit per day and divided by two (for
average).
S(x) =

5x
2

· x · 10
Total cost T(x) of delivery and storage during two consecutive delivery is
T(x) = 5000 +

5x
2

· x · 10

Average daily cost C(x) during the two consecutive delivery is
C(x) =
5000
x
+ 25x, x 0
When x → 0 or x → ∞, the average daily cost becomes large, hence the
minimum value of ‘x’ should exists in between these two limits ie 0 x ∞.
Now to get the critical point, differentiate cost relation with respect to ‘x’.
d C(x)
dx
= −
5000
x2
+ 25
On solving, two critical points are x = ±
√
200 ≈ ±14. Only x =
√
200 ≈ 14
lies in the domain of ‘x’. To verify the minimum cost, second derivative of
average day cost at x = 14 is

d2
C(x)
dx2

14
=

10000
x3

14
= +ve
This means, cabinet maker should ask for delivery of wood in period of 14
days. The average daily cost in rupees is
C(
√
200) =
5000
√
200
+ 25 ×
√
200 ≈ 707
The quantity of wood should schedule a delivery is Q = 5 · 14 = 70 units.
Solved Problem 1.26 Prove that the maximum volume of a cone that is
inscribed in a sphere of radius R is 8
27
times of the volume of sphere.
Solution Let a right circular cone is constructed inside a sphere whose
radius is R. Base radius of right circular cone is r and height is h. Semi-
vertical angle of cone is θ (say).

36 Maxima Minima
r
h
l
l
2
R
R
b
O
A
B C
D
E
θ
Figure 1.7: Inscribing a right circular cone in a sphere of radius R.
Now volume of cone is given by
V =
1
3
πr2
h (1.1)
Both r and h are variables. Converting radius and height of cone if form of
radius of sphere. From figure
sin θ =
l
2
R
That gives
l = 2R sin θ (1.2)
Again
sin θ =
r
l
Here l is slant height of the right circular cone.
cos θ =
h
l

Substituting the values of r and l in equation (1.1)
V =
1
3
πl2
sin2
θ l cos θ (1.3)
Substituting the value of l in above relation from equation (1.2), volume of
cone is
V =
8
3
πR3
cos4
θ sin2
θ (1.4)
To find the maximum volume differentiate to the equation (1.4) with respect
to θ and equation it to zero.
d
dθ
V =
d
dθ

π
8
3
R3
cos4
θ sin2
θ

Applying product rule for differentiation in right hand side of above equation.
Hence
d
dθ
V = π
8
3
R3

cos4
θ 2 sin θ cos θ − 4 cos3
θ sin θ sin2
θ

= 0
On simplification
2 cos2
θ = 4 sin2
θ
This gives
tan θ =
1
√
2
(1.5)
Applying right angle triangle rule
sin θ =
1
√
3
And
cos θ =
√
2
√
3
Now for maximum volume second derivative test for volume of cone must be
negative for the value of sin θ and cos θ. Now second derivative value of cone
volume is
d2
dθ2
V =
d2
dθ2

π
8
3
R3
cos4
θ sin2
θ

38 Maxima Minima
Second derivative is
d2
dθ2
V =
8
3
πR3

12 cos2
θ sin4
θ − 22 cos4
θ sin2
θ + 2 cos6
θ

Substituting the values of cos θ and sin θ in above relation
d2
dθ2
V =
8
3
πR3

24
27
−
88
27
+
16
27

= −ve (1.6)
This means volume of cone is maximum for given values of cos θ and sin θ.
Now Substituting the values of cos θ and sin θ in relation (1.1)
V =
8
3
πR3 4
9
1
3
V =
8
27
πR3 4
3
| {z }
Volume of Sphere
(1.7)
This gives that maximum volume of cone is 8
27
times of the volume of sphere.
Solved Problem 1.27 Find the maximum area of rectangle that is inscribed
in ellipse x2
a2 + y2
a2 = 1.
Solution A rectangle is drawn inside the ellipse as its four corners are at
the ellipse. Assume a point A whose coordinate is (x, y). From symmetry
other three coordinates of corners of rectangle are B(x, −y), C(−x, −y) and
D(−x, y). Now length of side AB of rectangle is 2y and length of side BC
is 2x. Area of rectangle is

x
y
O
A
B
C
D
(x, y)
(x, −y)
(−x, −y)
(−x, y) b
b
b
b
2y
2x
Figure 1.8: Inscribing a rectangle in a ellipse.
A = 2x × 2y (1.8)
From the definition of ellipse x2
a2 + y2
a2 = 1
y =
b
a
√
a2 − x2 (1.9)
Now the area of rectangle is
A = 4x
b
a
√
a2 − x2 (1.10)
For maximum or minimum area of rectangle, first derivative of equation
(1.10) with respect to x would be make equal to zero to find the extrema
points. ie
dA
dx
= 4 ·
b
a
·

√
a2 − x2 −
x2
√
a2 − x2

For stationary values of x equating above relation to zero and solving for x.
The result is
a2
− x2
= x2
x = ±
a
√
2

40 Maxima Minima
If are of rectangle is extremum then second derivative test of area A should
be negative
d2
A
dx2
= 4 ·
b
a
·

−
3 x
√
a2 − x2
−
x3
(a2 − x2)
3
2
#
(1.11)
On substituting the value of x, d2A
dx2 is negative, hence area of the rectangle
inscribed inside the ellipse is maximum when
x = ±
a
√
2
Now maximum area of inscribed rectangle is
A = 4 ·
b
a
·
a
√
2
s
a2 −

a
√
2
2
(1.12)
On solving equation (1.12), maximum area of rectangle is
A = 2ab (1.13)
Solved Problem 1.28 Find the maximum area of isosceles triangle that is
inscribed in ellipse x2
a2 + y2
a2 = 1.
Solution A isosceles triangle is drawn inside the ellipse as its tree vertices
are at the ellipse. Assume a point A whose coordinate is (a, 0). From sym-
metry of isosceles triangle other three coordinates are B(−x, −y), C(−x, y)
and D(−x, 0). Now length of height AD of triangle is (x + a) and length of
base BC of triangle is 2y. Area of triangle is

x
y
b
b
b
b
O A
B
C
D
(a, 0)
(−x, −y)
(−x, y)
(−x, 0)
a + x
2y
Figure 1.9: Inscribing a isosceles triangle in a ellipse.
A =
1
2
2y × (x + a) (1.14)
From the definition of ellipse x2
a2 + y2
a2 = 1
y =
b
a
√
a2 − x2 (1.15)
Now the area of triangle is
A = (x + a) ×
b
a
√
a2 − x2 (1.16)
To find maximum or minimum area of triangle, first derivative of equation
(1.16) with respect to x would be make equal to zero for extrema points. ie
dA
dx
=
b
a
·

√
a2 − x2 −
x(x + a)
√
a2 − x2

For stationary values of x equating above relation to zero and solving for x.
The result is
2x2
+ ax − a2
= 0
Solving it by using Shridharacharya formula (quadratic solution method)
x =
a
2
, −a

Maxima & Minima

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