Benginning Calculus Lecture notes 10 - max, min

Beginning Calculus
Applications of Di¤erentiation
- Maxima and Minima Problems -
Shahrizal Shamsuddin Norashiqin Mohd Idrus
Department of Mathematics,
FSMT - UPSI
(LECTURE SLIDES SERIES)
VillaRINO DoMath, FSMT-UPSI
(DA2) Applications of Di¤erentiation - Maxima and Minima Problems 1 / 12

Maximum and Minimum Problems
Learning Outcomes
Finding the critical points, critical values and end points.
Determine the optimal solutions.

Finding Maximum and Minimum Graphically
Max
Min
Finding the maximum and minimum of a graph of a function is easy
when the graph is sketched, but sketching the graph is time
consuming.

Keys to Finding Max/Min
Critical points
End Points
Points of discontinuity.

Example
Problem: A wire of length 1 is cut into two pieces. Each piece encloses
a square. Find the largest area enclosed.
Draw a diagram and name the variables.
1 unit length
x 1 - x
x/4 (1 –x)/4
Area:
A (x) =
x
4
2
+
1 x
4
2

Example - continue
Find the critical points, set A0 = 0 :
A0
(x) =
x
8
1 x
8
= 0 , x =
1
2
The critical value:
A
1
2
=
1
64
+
1
64
=
1
32
The endpoints, between 0 < x < 1.
lim
x!0+
A (x) =
1
16
, lim
x!1
A (x) =
1
16

Example - continue
Minimum area enclosed is A
1
2
=
1
32
(these are equal squares) -
the minimum value.
Maximum area enclosed is A (0) =
1
16
or A (1) =
1
16
- the
maximum values.
The minimum value occurs at x =
1
2
, and the maximum values
occurs at x near 0 or x near 1.
Alternatively, the minimum point is
1
2
,
1
32
, and the maximum
points are 0,
1
16
and 1,
1
16
.

Example
Problem: Find a box (with square bottom) without a top with least
surface area for a …xed volume.
Draw the diagram and name the variables:
x
x
y
Volume:
V = x2
y, y …xed, volume …xed
Surface Areas:
A = 4xy + x2

Example - continue
y =
V
x2
. Then,
A (x) = 4
V
x
+ x2
The critical point(s):
A0
(x) =
4V
x2
+ 2x = 0 , x = (2V )1/3
The end point(s), between 0 < x < ∞ :
lim
x!0+
A (x) = ∞, lim
x!∞
A (x) = ∞
Using the second derivative (to test the concavities)
A00
(x) =
8V
x3
+ 2 > 0 for 0 < x < ∞ concave up

Example - continue
The least surface area occurs at x = 21/3V 1/3 and y = 2 2/3V 1/3
The surface area is
A (x) = 4
V
21/3V 1/3
+ 21/3
V 1/3
2
= 3 21/3
V 2/3

Example - continue - More meaningful answer
Dimesionless variable
A
V 2/3
= 3 21/3
The ratio:
x
y
=
21/3V 1/3
2 2/3V 1/3
= 2
the length x is twice the height. The optimal shape of the box.

Example - continue - Using implicit di¤erentiation
V = x2y and A = 4xy + x2
d
dx
(V ) =
d
dx
x2
y
0 = 2xy + x2
y0
) y0
=
2y
x
d
dx
(A) =
d
dx
4xy + x2
0 = 4y + 4xy0
+ 2x = 4y + 4x
2y
x
+ 2x = 2x 4y
)
x
y
= 2
Advantage: nicer and faster than previous method.
Disadvantage: did not check whether the critical point(s).

Benginning Calculus Lecture notes 10 - max, min

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Benginning Calculus Lecture notes 10 - max, min