Application of Derivatives — Class 12 Maths Solution

Let $x{\rm{cm}}$ be the length of each side of the square which is to be cut off from each corner of the square tin sheet of side $18{\rm{ cm}}$.

Let $V$ be the volume of the open box formed by folding up the flaps, then
$V = x(18 - 2x)(18 - 2x\rangle = 4x{(9 - x)^2}$
$= 4({x^3} - 18{x^2} + 81x)$ …(i)

Differentiating (i) w.r.t. $x$, we get

$\cfrac{{dV}}{{dx}} = 4(3{x^2} - 36x + 81) = 12({x^2} - 12x + 27)$
For maximum/minimum volume,
$\cfrac{{dV}}{{dx}} = 0 \Rightarrow 12({x^2} - 12x + 27) = 0$

$\Rightarrow 12(x - 3)(x - 9) = 0 \Rightarrow x = 3,9$ but $0 < x < 9 \Rightarrow x = 3$

$\cfrac{{{d^2}V}}{{d{x^2}}} = 12(2x - 12) = 24(x - 6)$
and ${\left( {\cfrac{{{d^2}V}}{{d{x^2}}}} \right)_{x = 3}} = 24(3 - 6) = - 72 < 0$

$\Rightarrow V$ has a maximum value at $x = 3$.

Hence we can say that the volume of the box is maximum when the side of the square to be cut off is $3{\rm{ cm}}$.