Question

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 3131 in. by 1717 in. by cutting congruent squares from the corners and folding up the sides. Then find the volume.

Answer 1

840.02 square inches ( approx )

Explanation:

Suppose x represents the side of each square, cut from the corners of the sheet,

Since, the dimension of the sheet are,

31 in × 17 in,

Thus, the dimension of the rectangular box must are,

(31-2x) in × (17-2x) in × x in

Hence, the volume of the box would be,

V = (31-2x) × (17-2x) × x

`=(31* 17 +31* -2x -2x* 17 -2x* -2x)x`

`=(527 -62x-34x+4x^2)x`

`\implies V=4x^3-96x^2 +527x`

Differentiating with respect to x,

`(dV)/(dx)=12x^2-192x+527`

Again differentiating with respect to x,

`(d^2V)/(dx^2)=24x-192`

For maxima or minima,

`(dV)/(dx)=0`

`\implies 12x^2-192x+527=0`

By the quadratic formula,

`x=(192 \pm √(192^2 -4* 12* 527))/(24)`

`x\approx 8\pm 4.4814`

`\implies x\approx 12.48\text{ or }x\approx 3.52`

Since, at x = 12.48,
`(d^2V)/(dx^2)` = Positive,

While at x = 3.52,
`(d^2V)/(dx^2)` = Negative,

Hence, for x = 3.52 the volume of the rectangle is maximum,

Therefore, the maximum volume would be,

V(3.5) = (31-7.04) × (17-7.04) × 3.52 = 840.018432 ≈ 840.02 square inches