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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
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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