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

Answer:

840.02 square inches ( approx )

Step-by-step 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

$=(31\times 17 +31\times -2x -2x\times 17 -2x\times -2x)x$

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

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

Differentiating with respect to x,

$\frac{dV}{dx}=12x^2-192x+527$

Again differentiating with respect to x,

$\frac{d^2V}{dx^2}=24x-192$

For maxima or minima,

$\frac{dV}{dx}=0$

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

By the quadratic formula,

$x=\frac{192 \pm \sqrt{192^2 -4\times 12\times 527}}{24}$

$x\approx 8\pm 4.4814$

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

Since, at x = 12.48, $\frac{d^2V}{dx^2}$ = Positive,

While at x = 3.52, $\frac{d^2V}{dx^2}$ = 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