Dimensions of Rectangular Sheet Yielding Maximum Box Volume

To find the side lengths of the original rectangular sheet, we proceed step-by-step:

Step 1: Set up the dimensions of the sheet and the cut-outs Let the dimensions of the rectangular sheet be $a$ and $b$. Given that their lengths are in the ratio $8 : 15$, we can write: $a = 8k \quad \text{and} \quad b = 15k \quad (k > 0)$

Squares of side length $x$ are removed from each of the four corners. The total area of the four removed squares is given as $100$: $4x^2 = 100 \implies x^2 = 25 \implies x = 5$

Step 2: Express the volume of the box as a function of $x$ After cutting out the corners and folding up the flaps, the dimensions of the resulting open rectangular box are:

• Height: $x$
• Width: $a - 2x$
• Length: $b - 2x$

The volume $V(x)$ of the box is given by: $V(x) = x(a - 2x)(b - 2x) = 4x^3 - 2(a + b)x^2 + abx$

Step 3: Apply the condition for maximum volume For a given sheet, the volume is maximized when the first derivative of $V(x)$ with respect to $x$ is equal to zero, i.e., $V'(x) = 0$: $V'(x) = 12x^2 - 4(a + b)x + ab = 0$

Since the maximum volume is attained when the removed square area is $100$ (which corresponds to $x = 5$), $x = 5$ must satisfy the equation: $12(5)^2 - 4(a + b)(5) + ab = 0$ $300 - 20(a + b) + ab = 0$

Substitute $a = 8k$ and $b = 15k$: $300 - 20(8k + 15k) + (8k)(15k) = 0$ $300 - 20(23k) + 120k^2 = 0$ $120k^2 - 460k + 300 = 0$

Dividing through by $20$: $6k^2 - 23k + 15 = 0$

Step 4: Solve for $k$ Factoring the quadratic equation: $6k^2 - 18k - 5k + 15 = 0$ $6k(k - 3) - 5(k - 3) = 0$ $(6k - 5)(k - 3) = 0$

This yields two roots: $k = 3 \quad \text{or} \quad k = \frac{5}{6}$

• If $k = \frac{5}{6}$, then $a = 8\left(\frac{5}{6}\right) = \frac{20}{3} \approx 6.67$. For this value, the cut-out length $2x = 10 > a$, which is physically impossible as the width $a - 2x$ would be negative.
• If $k = 3$, then: $a = 8(3) = 24$ $b = 15(3) = 45$ Here, $a - 2x = 24 - 10 = 14 > 0$ and $b - 2x = 45 - 10 = 35 > 0$, which is physically valid.

Furthermore, checking the second derivative: $V''(5) = 24(5) - 4(24 + 45) = 120 - 276 = -156 < 0$ This confirms that the volume is indeed maximized at $x = 5$.

Conclusion: The side lengths of the original rectangular sheet are $24$ and $45$.

Therefore, the correct options are A and C.