I have a square piece of cardboard that is 24 inches on each side. i want to cut out a square in each corner and the fold up the sides to make an open box. how big should i make the squares that i cut out in order to make the box have the greatest volume?

A square of side 4 inch must be cut in order to get greatest volume.

after cutting a square of side x inch and folding it into a box, the length of each side of the box's base is (24-2x) inch and box's height is x inch.
∴volume, V= (24-2x)² × x
differentiate V with respect to x and put dV/dx = 0
we get x =12 and x= 4
x=12 is not possible so maximum volume is obtained when x=4 inch.

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MrRoyal MrRoyal · Answer 2 · 2021-10-19T11:54:48+02:00

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

The square to cut out should be 8 inches long in order to produce the greatest volume

The side length of the square is:

[tex]Length= 24[/tex]

Assume the cut-out is x

The dimension of the box would be

[tex]Length=Width= 24-2x[/tex]

[tex]Height = x[/tex]

The volume of the box is:

[tex]V= Length \times Width \times Height[/tex]

Substitute known values

[tex]V = (24 - 2x) \times (24 - 2x) \times x[/tex]

Expand

[tex]V = (24 - 2x) \times (24x - 2x^2)[/tex]

Expand

[tex]V = 576x - 48x^2 - 48x^2 + 4x^3[/tex]

Differentiate

[tex]V' = 576 - 96x - 96x + 12x^2[/tex]

[tex]V' = 576 -192x + 12x^2[/tex]

Set to 0

[tex]576 -192x + 12x^2 = 0[/tex]

Divide through by 12

[tex]48 -16x + x^2 = 0[/tex]

Rewrite as:

[tex]x^2 - 16x + 48 = 0[/tex]

Expand

[tex]x^2 - 8x -8x + 48 = 0[/tex]

Factorize

[tex]x(x - 8) -8(x - 8) = 0[/tex]

Factor out x - 8

[tex](x - 8)(x - 8) = 0[/tex]

Split

[tex]x - 8 = 0[/tex] or [tex]x - 8 = 0[/tex]

Solve for x

[tex]x = 8[/tex]

Hence, the square to cut out should be 8 inches long in order to produce the greatest volume

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