Question

An open box with a square base is constructed so that it has a volume of 64in^3. The base needs to be made out of a material that costs twice as much as the sides. What are the dimensions of the box, what are the dimensions of the box x and y (x being the width and depth of the square dimensions, and y being the height of the box) which will minimize the cost of the box?

Answer 1

The volume of a box is the amount of space in the box

The dimensions that minimize the cost of the box is 4 in by 4 in  by 4 in

How to determine the dimensions that minimize the cost

The dimensions of the box are:

Width = x

Depth = y

So, the volume (V) is:

$V = x^2y$

The volume is given as 64 cubic inches.

So, we have:

$x^2y = 64$

Make y the subject

$y = \frac{64}{x^2}$

The surface area of the box is calculated as:

$A =x^2 + 4xy$

The cost is:

$C = 2x^2 + 4xy$ --- the base is twice as expensive as the sides

Substitute $y = \frac{64}{x^2}$

$C = 2x^2 + 4x * \frac{64}{x^2}$

$C =2x^2 + \frac{256}{x}$

Differentiate

$C' =4x - \frac{256}{x^2}$

Set to 0

$4x - \frac{256}{x^2} = 0$

Multiply through by x^2

$4x^3 - 256= 0$

Divide through by 4

$x^3 - 64= 0$

Add 64 to both sides

$x^3 = 64$

Take the cube roots of both sides

$x = 4$

Recall that:

$y = \frac{64}{x^2}$

So, we have:

$y = \frac{64}{4^2}$

$y = 4$

Hence, the dimensions that minimize the cost of the box is 4 in by 4 in  by 4 in

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