# From a thin piece of cardboard 30 in. by 30 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum...

From a thin piece of cardboard 30 in. by 30 in., square corners are cut out so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? What is the maximum volume? Round to the nearest tenth, if necessary.

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Let h be the length (in inches) of the cutout portion of the square corners. Eventually, that would be the height of the cardboard box.

Both sides of the cardboard would reduce by 2h, owing to the cut.

The box is thus (30 – 2h) by (30 – 2h) by h in dimensions.

The volume V (in cubic inches) is:

`V = h(30 -2h)^2`

`=h(900-2*30*2h+4h^2)`

`=4h^3-120h^2+900h`

For obtaining a box of maximum volume, maximize V as a function of h.

Differentiate both sides with respect to h,

`(dV)/(dh)=4*3*h^2-120*2*h+900`

`=12h^2-240h+900`

`=12(h^2-20h+75)`

`=12(h^2-15h-5h+75)`

`=12(h(h-15)-5(h-15))`

`=12(h-15)(h-5)`

For maximum, (or for that matter, minimum), `(dV)/(dh)=0`

`So, 12(h-15)(h-5)=0`

Put each term equal to zero to obtain:

h=15, and 5

When `h=15, V= 15(30 -2*15)^2=0`

This corresponds to the minimum volume.

When` h=5, V= 5(30 -2*5)^2=5*20^2`

`=2000 ` cubic inches.

This is the maximum volume the box can assume.

Dimensions of the box to yield a maximum volume is therefore,

20 by 20 by 5 in.

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