A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed square is 100, the resulting box has maximum value, the dimensions of the sides of the rectangular sheet are 

24, 45

Let the sides of the rectangular sheet be 15a and 8a units and the length of the side of each square to be cut from each corner of the sheet be x units. Then the dimensions of the box are

Length = $15 a-2 x$, Breadth = $8 a-2 x$, Depth = x

Clearly, Length > 0, Breadth > 0 and Depth > 0. Therefore, 0 < x < 4a.

Let V be the volume of the box. Then,

$V=(15 a-2 x)(8 a-2 x) x=120 a^2 x-46 a x^2+4 x^3$

∴  $\frac{d V}{d x}=120 a^2-92 a x+12 x^2$ and $\frac{d^2 V}{d x^2}=-92 a+24 x$

The critical numbers of V are given by $\frac{d V}{d x}=0$

∴  $\frac{d V}{d x}=0$

$\Rightarrow 120 a^2-92 a x+12 x^2=0$

$\Rightarrow 30 a^2-23 a x+3 x^2=0$

$\Rightarrow (6 a-x)(5 a-3 x)=0$

$\Rightarrow 5 a-3 x=0$                    [∵ 0 < x < 4a    ∴ 6a - x ≠ 0]

$\Rightarrow x=\frac{5 a}{3}$

When $x=\frac{5 a}{3} \frac{d^2 V}{d x^2}=-92 a+40 a=-52 a<0$

It is given that the total area of all squares cut from each corner of the sheet is 100 sq. units.

Therefore, $4 x^2=100 \Rightarrow 4\left(\frac{5 a}{3}\right)^2=100 \Rightarrow a^2=9 \Rightarrow a=3$

Hence, the dimensions of the sheet are 8a = 24 and 15a = 45 units.