Find the dimensions that will make the cost a minimum.

Question

The base of a rectangular box is to be twice as long as it is wide. The volume of the box is 500 cubic inches. The material for the top costs $0.50 per square inch and the material for the sides and bottom costs $0.25 per square inch. Find the dimensions that will make the cost a minimum.

length	in
width	in
height	in

Answer 1

L=2W

LWH = 500

2WWH = 500

W^2H = 250

H=250/W^2

C=.5LW + .25LW + 2(.25LH) + 2(.25WH)

C=.5(2W)W + .25(2W)W + 2(.25(2W)(250/W^2) + 2(.25)W(250/W^2)

C= W^2 +.5W^2 + 250/W + 125/W

= 1.5W^2 + 375/W

C'(W) = 3W -375/W^2 = 0

3W^3 = 375

W^3 = 125

W = 5 in. wide

L= 10 in. long

H= 250/25 = 10 in. high

minimizes cost

Answer 2

ANSWER;

Length = 10 inches

Width = 5 inches

Height = 10 inches

STEP-BY-STEP EXPLANATION;

If the length, width, and height are L, W, and H, then:

L = 2W

500 = LWH

If the cost is C then;

C = 0.50LW + 0.25 (LW + 2LH + 2WH)

C = 0.50LW + 0.25LW + 0.50LH + 0.50WH

C = 0.75LW + 0.50LH + 0.50WH

We need the cost in terms of a single variable.

Substitute the first equation into the second equation to get H in terms of W;

500 = (2W) WH

500 = 2W²H

250 = W²H

H = 250/W²

Now substitute the equations for L and H into the cost equation;

C = 0.75(2W)W + 0.50(2W)(250/W²) + 0.50W(250/W²)

C = 1.5W² + 250/W + 125//W

C = 1.5W² + 375/W

Take derivative and set to 0;

dC/dW = 3W - 375/W²

0 = 3W - 375/W²

3W = 375/W²

3W³ = 375

W³ = 125

W = 5

Therefore;

L = 10

Substitute to get H;

500 = LWH

500 = 10 * 5 * H

500 = 50H

H = 10