Question

If you want to produce an open-top box from a 16 in by 22 in flat piece of cardboard, what is the maximum possible volume (in in3) of the box? Round your answer to the nearest whole number and omit units.

Answer 1

480

Explanation:

Since, for making a box from a cardboard,

We need to cut four congruent pieces from each corner of the cardboard,

Let x be the side of a piece ( in inches ),

Given,

The dimensions of the cardboard are 16 in by 22,

So, the dimension of the box would be (16-2x) in by (22-2x) in by x in,

Thus, the volume of the box,

`V(x)=(16-2x)(22-2x)x=4x^3-76x^2+352x`

Differentiating with respect to x,

`V'(x) = 12x^2-152x+352`

Again differentiating with respect to x,

`V''(x) = 24x-152`

For maxima or minima,

V'(x) = 0

`\implies 12x^2-152x+352=0`

By the quadratic formula,

`x=(-(-152)\pm √(-152^2-4* 12* 352))/(24)`

`x=(152\pm √(6208))/(24)`

`\implies x\approx 9.62 \text{ or }x\approx 3.05`

Since, for x = 9.62, V''(x) = positive,

While for x = 3.05, v''(x) = negative,

Hence, volume is maximum for x = 3.05,

And, maximum volume,

`V(3.05) = 4(3.05)^3-76(3.05)^2+352(3.05)=480.1005\approx 480\text{ cube in}`