Question

Richard has a 9-inch by 13.5-inch sheet of cardboard. He cuts squares from 4 corners to create an open container with the largest possible volume.
What is the greatest volume possible? Express your answer as a decimal to the nearest cubic inch.

Answer 1

Answer:

96.256 in³

Step-by-step explanation:

Given,

The dimension of the sheet = 9-inch by 13.5-inch,

Suppose x be the side of the square which is cut from each corner,

So, the dimension of the resultant open container,

(9-2x) inch × (13.5 - 2x) inch × x inch

Thus, the volume of the box,

V(x) = (9-2x) × (13.5 - 2x) × x

$V(x) = (9x-2x^2)(13.5 - 2x)$

Differentiating with respect to x,

$V'(x) = (9x - 2x^2)(-2) + (9 - 4x)(13.5 - 2x)$

$=-18x + 4x^2 + 121.5 - 18x - 54x + 8x^2$

$=12x^2 - 90x + 121.5$

Again differentiating,

$V''(x) = 24x - 90$

For maxima or minima,

$V'(X) = 0$

$12x^2 - 90x + 121.5=0$

Using the quadratic formula,

$x =\frac{90\pm \sqrt{90^2 - 4\times 12\times 121.5}}{24}$

$\implies x\approx 1.766\text{ or }x=5.734$

For x = 1.766, V''(x) = negative,

Thus, V(x) is maximum at x = 1.766,

For x =5.734, V''(x) = positive,

Thus, V(x) is minimum at x = 5.734,

Hence, the maximum possible volume, V(1.766) = (9-2(1.766)) × (13.5 - 2(1.766)) × 1.766

≈ 96.256 in³