An open box with a square base is to have a volume of 6 ft^3

Question

(a) Find a function that models the surface area A of the box in terms of the length of one side of the base x.

(b) Find the box dimensions that minimize the amount of material used. (Round your answers to two decimal places.)

height:

side length:

Thanks!

Answer 1

a) Volume V = x²h, where x is side of square base and h is hite.

Then surface area S = x² + 4xh because box is open.

b) From V = x²h = 6 we have h = 6/x².

Substitude in formula for surface area: S = x² + 4x·6/x², S = x² + 24/x.

We get S as function of one variable x. To get minimum we have to find derivative S' = 2x - 24/x² = 0,

from here 2x³ - 24 = 0, x³ = 12, x = (12)^1/3 ≅ 2.29 ft.

Then h = 6/(12)^2/3 = (12)^1/3/2 ≅ 1.14 ft.

To prove that we have minimum let get second derivative: S'' = 2 + 48/x³, S''(12^1/3) = 2 + 48/12 = 6 > 0.

And because by second derivative test we have minimum: S_min = (12)^2/3 + 4(12)^1/3(12)^1/3/2 = 3(12)^2/3 ≅ 15.72 ft²