Applied max min
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1. The largest rectangular area that can be enclosed using 100 feet of fencing against a long straight wall is 1250 square feet. The dimensions are 50 feet by 50 feet. 2. The dimensions of a rectangular area of 3200 square feet that minimizes fencing costs are 40 feet by 80 feet. 3. The dimensions of a square-bottomed box with a fixed volume of 500 cubic centimeters that minimizes surface area are 7.94 centimeters by 7.94 centimeters by 7.94 centimeters.
![a) Amount of fencing available
Area constraints
2x2 ≥800 so x ≥ 20
y2 ≥100 so y ≥ 10
150 -1.5x ≥10
x≤ 140/1.5
x≤ 932/3
Domain of x
[20, 932/3]
Graph of A(x) has
A (932/3) = 17,522.2
A(20) = 15,200
no local max so
abs max value of A occurs
at an endpoint of
the domain. Max area enclosed is
17522. 2 ft2.](https://image.slidesharecdn.com/appliedmaxmin-111128080627-phpapp01/85/Applied-max-min-21-320.jpg)
Applied max min
- 1. What are the dimensions of the box with largest volume which can be made by cutting squares from the corners of an 8.5 "x 11" piece of cardboard and turning up the edges?
- 2. Find the dimensions of the rectangle of largest area that has its base on the x axis and its other two vertices above the x axis and lying on the curve f(x) = .
- 3. Which point on the graph of y = ex is closest to the origin?
- 4. 1. If you have 100 feet of fencing and you want to enclose a rectangular area up against a long, straight wall, what is the largest area you can enclose? y y x 100 = 2y + x A = xy A = x(100 -x) 2 A(x) = 100x -x2 2 A'(x) = 50 - x Max/Min Area when A'(x) = 0 i.e. when x = 50. A'(x) + 0 0 50 Max area occurs when x = 50 and the maximum area is 50(50)/2 or 1250 sq feet
- 5. 2 2. A rectangular area of 3200 ft is to be fenced off. Two opposite sides will use fencing costing $2 per foot and the remaining sides will use fencing costing$1 per foot. Find the dimensions of least cost. LW = 3200 W = 3200 W L $1 / ft L $2 / ft Cost = length x Cost per unit of length Cost = 2L ( 2) + 2W (1) C(L) = 4L + 2 3200 L Max/Min cost will occur when C'(L) = 0 0 C'(L) C'(L) = 4 - 6400 L2 0 40 C(L) C'(L) = 0 = 4 - 6400 L2 When L = 40, W = 3200/40 = 80 so the cost is minimized if the L2 = 1600 rectangular area has dimensions 40 ft x 80 ft. L = 40
- 6. 3. A square-bottomed box with a top has a fixed volume of 500 cm 3. What dimensions minimize the surface area? V = 500 = x2h h 500 = h x2 Surface area S = 2x2 +4xh x x = 2x2 +4x 500 x2 S = 2x2 + 2000 Minimize x S'(x) = 4x - 2000 x2 0 = 4x - 2000 x2 0 = 4(x3 - 500) x2 S'(x) 0 x = ∛500 = 7.94 cm S(x) 7.94 when x = 7.94, the surface area is minimized and h = 500/ 7.942 = 7.94 Dimensions of box which minimizes surface area are 7.94 cm x 7.94 cm x 7.94 cm
- 7. 4. A square-bottomed box without a top has fixed volume of 500 cm3. What dimensions minimize the surface area? V = 500 = x2h h 500 = h x2 Surface area S = x2 +4xh x x = x2 + 4x 500 x2 S = x2 + 2000 Minimize x S'(x) = 2x - 2000 x2 0 = 2x - 2000 x2 S'(x) 0 3 0 = 2(x - 1000) x2 S(x) 10 x = ∛1000 = 10 cm when x = 10, the surface area is minimized and h = 500/ 102 = 5 Dimensions of box which minimizes surface area are 10 cm x 10 cm x 5 cm
- 8. (x, y) (x, y) L y L 1-x (1,0) L2 = (1- x)2 + y2 b ut y = x2 so L2 = (1 - x)2 + x4 minimize this 2LdL = 2(1 - x)(-1) + 4x3 dL dx dx = (1 - x)(-1) + dL 3 2x 0 = -1 + x + 2x3 dx = x = .589 (1-x)2 + x4 L'(x) 0 L(x) .589 Distance from (x,y) is a minimum when x =.589 and y = .5892 = .347 So, the point on the parabola y = x2 which is closest to the point (1, 0) is (.589, .347).
- 9. (x,y) L2 = (1 - x)2 + (y - 2)2 L = (1 - x)2 + ( 4 - x2 - 2)2 = (1 - x)2 + (2 - x2 )2 (1,2) N eed to find the minimum value of L L2 = (1 - x)2 + (2 - x2)2 2 L L'(x) = 2(1 - x)(-1) + 2(2 - x2)(-2x) L'(x) = -2 + 2x - 8x + 4x3 2 √ ((1 - x)2 + (2 - x2 )2) L'(x) = -1 - 3x + 2x3 √ (1 - x)2 + (2 - x2 )2 L'(x) = 0 when 0 = -1 - 3x + 2x3 x = 1.366 if 0 < x < 2 Since L' (x) < 0 for x < 1.366 and L'(x) > 0 for x > 1.366, L has a minimum value when x = 1.366. When x = 1. 366, y = 4 - 1.3662 = 2.134 So (1.366, 2.134) is the point on the curve y = 4 - x2 which is closest to the point (1,2)
- 10. SA = 210 = 2πr2 + 2πrh h = 210 - 2πr2 h 2πr r V = πr2h V(r) = πr2 210 - 2πr2 2πr V(r) = r(210 - 2πr2) Maximize 2 V'(r) = (210 - 6πr2) 2 V'(r) = 0 when 210 - 6πr2 = 0 210 = 6πr2 V'(r) 0 105 = r2 0 3.34 3π √(35/π) = r = 3.34 cm The volume is a maximum when r = 3.34 cm and h = (210 - 2π(3.34)2 )/(2π(3.34)) = 6.68 cm
- 11. Minimum surface area when r = 5.42 cm and h = 500/(π x 5.422) = 5.42 cm.
- 14. 11. A rectangle is bounded by the x and y axes and the line f(x) = 6 - x 2 What length and width should the rectangle have so that its area is a maximum?
- 15. 12. A right triangle is formed in the first quadrant by the x and y axes and a line through the point (1,2). Find the vertices of the triangle so that its area is a minimum.
- 16. 13. Find the dimensions of the rectangle of largest area that has its base on the x axis and its other two vertices above the x axis and lying on the parabola y = 8 - x2 . (x, y) y = 8 - x2
- 17. 14. A rectangle is drawn with sides parallel to the coordinate axes with its upper two vertices on the parabola f(x) = 4 - x2 and its lower two vertices on the parabola g(x) = x2/2 -2. What is the maximum possible area of the rectangle? (x, f(x)) (x, g(x))
- 18. 15
- 19. 16.
- 21. a) Amount of fencing available Area constraints 2x2 ≥800 so x ≥ 20 y2 ≥100 so y ≥ 10 150 -1.5x ≥10 x≤ 140/1.5 x≤ 932/3 Domain of x [20, 932/3] Graph of A(x) has A (932/3) = 17,522.2 A(20) = 15,200 no local max so abs max value of A occurs at an endpoint of the domain. Max area enclosed is 17522. 2 ft2.