1. Let the sides of the rectangular sheet be in the ratio $8:15$.

   So, let $L=8k, \qquad B=15k$ for some positive constant $k$.

2. Let squares of side $x$ be removed from each corner.

   Since 4 equal squares are removed and the total removed area is $100$, we have $4x^2=100$ $x^2=25$ $x=5$ (taking positive value since length is positive).

3. Volume of the open box formed.

   After folding, the dimensions of the box are:

   • length $=8k-2x=8k-10$
   • breadth $=15k-2x=15k-10$
   • height $=x=5$

   Hence, $V=5(8k-10)(15k-10)$

4. Use the condition that the box has maximum volume.

   For a rectangle of fixed perimeter, if squares of side $x$ are cut and folded, volume is $V=x(l-2x)(b-2x)$ and for fixed $l+b$, maximum volume occurs when $l+b=6x$ because $V=x(l-2x)(b-2x)=x(lb-2x(l+b)+4x^2)$ and differentiating w.r.t. $x$ for fixed $l+b,lb$ gives $\frac{dV}{dx}=lb-4x(l+b)+12x^2$ At maximum, with given ratio fixed, it is easier to use the standard condition directly through fixed perimeter.

   Since here $x=5$, we get $l+b=6x=30$

5. Now use the side ratio $8:15$.

   Let sides be $8m$ and $15m$. Then $8m+15m=30$ $23m=30$ $m=\frac{30}{23}$

   This does not match the options, so let us instead maximize directly using ratio parameter and fixed cut size $x=5$.

6. Direct maximization with ratio fixed.

   $V(k)=5(8k-10)(15k-10)$

   Expanding: $V(k)=5(120k^2-230k+100)=600k^2-1150k+500$

   This is a quadratic opening upward, so this expression alone cannot represent a maximum unless another condition is used: the sheet has fixed perimeter.

7. Interpret fixed perimeter correctly.

   Let original sheet sides be $l$ and $b$ with fixed perimeter, so $2(l+b)=P \Rightarrow l+b=\text{constant}$ and also $l:b=8:15$ Therefore both $l$ and $b$ are uniquely determined by the perimeter. So the variable for maximization is actually the cut size $x$.

   Hence volume is $V=x(8k-2x)(15k-2x)$ with $k$ fixed.

   Differentiate: $V=x(120k^2-46kx+4x^2)=120k^2x-46kx^2+4x^3$ $\frac{dV}{dx}=120k^2-92kx+12x^2$

   For maximum volume, set $120k^2-92kx+12x^2=0$

   Now $x=5$, so $120k^2-460k+300=0$ Divide by 20: $6k^2-23k+15=0$

   Factor: $6k^2-23k+15=(3k-5)(2k-3)=0$

   So, $k=\frac{5}{3} \quad \text{or} \quad k=\frac{3}{2}$

8. Find the corresponding sheet dimensions.

   Since sides are $8k$ and $15k$:

   • If $k=\frac{3}{2}$: $8k=12, \quad 15k=22.5$ These are not among the options.

   • If $k=\frac{5}{3}$: $8k=\frac{40}{3}, \quad 15k=25$ Again not among the options.

   This suggests we should parametrize as $l=16k, b=30k$ equivalent to ratio $8:15$ in option scale. Let us test option pairs directly.

9. Check option pairs having ratio $8:15$.

   From the options $24,32,45,60$, the only pair in ratio $8:15$ is $24:45=8:15$

10. Verify maximum condition with these dimensions.

Here $l=24, b=45$, and total removed area $100$ gives $x=5$.

Volume: $V(x)=x(24-2x)(45-2x)$

Differentiate: $V=x(1080-138x+4x^2)=1080x-138x^2+4x^3$ $V'(x)=1080-276x+12x^2$

At $x=5$: $V'(5)=1080-1380+300=0$

Also, $V''(x)=-276+24x$ $V''(5)=-276+120=-156<0$

So $x=5$ indeed gives maximum volume.

11. Therefore the sheet sides are $24 \text{ and } 45$

So the correct options are A and C.