Solve the problem.
-A company wishes to manufacture a box with a volume of 40 cubic feet that is open on top and is twice as long as it is wide. Find the width of the box that can be produced using the minimum amount of material. Round to
The nearest tenth, if necessary.
A) 7.2 ft
B) 3.6 ft
C) 3.2 ft
D) 6.4 ft

Question

Quizplus · Accepted Answer

Let the width of the box be x feet. Then the length of the box is 2x feet and the height is 20/x feet. The amount of material needed to make the box is the surface area, which is given by:

SA = 2lw + 2lh + 2wh
   = 2x(2x) + 2x(20/x) + 2(2x)(20/x)
   = 4x^2 + 40/x + 80/x
   = 4x^2 + 120/x

To minimize the amount of material, we need to find the value of x that minimizes the surface area. We can do this by taking the derivative of SA with respect to x, setting it equal to zero, and solving for x:

dSA/dx = 8x - 120/x^2
       = 8x(1 - 15/x^3)

Setting dSA/dx = 0, we get:

8x(1 - 15/x^3) = 0

Solving for x, we get:

x = 15^(1/3)

Substituting this value of x back into the equation for SA, we get:

SA = 4(15^(2/3)) + 120/(15^(1/3))
   ≈ 68.8

Rounding to the nearest tenth, we get the final answer of 3.2 ft. Therefore, the best choice is C. Let the length of the box be 2x and the width be x. Then the height must be 40/(2x^2), or 20/x^2.
The surface area of the box is given by A = 2x(2x) + 2x(20/x^2) = 4x^2 + 40/x.
To minimize the amount of material used, we need to minimize the surface area. Taking the derivative and setting it equal to 0, we get:
dA/dx = 8x - 40/x^2 = 0
Solving for x, we get x = 3.2.
Therefore, the width of the box that can be produced using the minimum amount of material is x = 3.2 ft.

Solve the Problem. -A Company Wishes to Manufacture a Box with a Volume