Question

By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, an open box may be made. If the cardboard is 15 in. long and 7 in. wide, find the dimensions (in inches) of the box that will yield the maximum volume. (Round your answers to two decimal places if necessary.)

Answer 1

Length = 12 in

Width = 4 in

Height = 1.5 in

Explanation:

Let the side length of the squares that are cut off be x.

By cutting the squares off, the height of the box formed will be x.

The length will be the length of cardboard - 2x =
`15 - 2x`.

The width will be the width of cardboard - 2x =
`7-2x`

The volume is given by

`V = x(15-2x)(7-2x) = 4x^3 -44x^2 +105x`

To maximize the volume,

`(dV)/(dx) = 0`

`(dV)/(dx) = 12x^2-88x+105=0`

`\left(2x-3\right)\left(6x-35\right) = 0`

`2x-3 = 0` OR
`6x-35 = 0`

`x = (3)/(2)` OR
`x=(35)/(6)`

To determine which value of x gives a maximum, we evaluate

`(d^2V)/(dx^2)`

`(d^2V)/(dx^2) = 24x-88`

The value of x that gives a negative value is maximum.

At
`x = (3)/(2)`,

`(d^2V)/(dx^2) = 24((3)/(2)) - 88 = 36-88 = -52`

We can confirm the other value will be positive:

`(d^2V)/(dx^2) = 24((35)/(6)) - 88 = 140-88 = 52`

The maximum occurs when
`x = (3)/(2)`.

This corresponds to

height =
`x = (3)/(2)`

length =
`15-2x = 15-2((3)/(2)) = 15 - 3 = 12`

width =
`7 - 2((3)/(2)) = 7-3 = 4`