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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?

Sagot :

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:

[tex]V = x^2y[/tex]

The volume is given as 64 cubic inches.

So, we have:

[tex]x^2y = 64[/tex]

Make y the subject

[tex]y = \frac{64}{x^2}[/tex]

The surface area of the box is calculated as:

[tex]A =x^2 + 4xy[/tex]

The cost is:

[tex]C = 2x^2 + 4xy[/tex] --- the base is twice as expensive as the sides

Substitute [tex]y = \frac{64}{x^2}[/tex]

[tex]C = 2x^2 + 4x * \frac{64}{x^2}[/tex]

[tex]C =2x^2 + \frac{256}{x}[/tex]

Differentiate

[tex]C' =4x - \frac{256}{x^2}[/tex]

Set to 0

[tex]4x - \frac{256}{x^2} = 0[/tex]

Multiply through by x^2

[tex]4x^3 - 256= 0[/tex]

Divide through by 4

[tex]x^3 - 64= 0[/tex]

Add 64 to both sides

[tex]x^3 = 64[/tex]

Take the cube roots of both sides

[tex]x = 4[/tex]

Recall that:

[tex]y = \frac{64}{x^2}[/tex]

So, we have:

[tex]y = \frac{64}{4^2}[/tex]

[tex]y = 4[/tex]

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

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