A rectangular sheet of tin $45{\rm{ cm by }}24{\rm{ cm}}$ is to be made into a box without top, by cutting off square from each corner and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
Let the side of the square which needs to be cut from each of the four corners be $x{\rm{cm}}$, then the base of the box has dimensions $(45 - 2x)$ cm and $(24 - 2x)$ cm and height of the box is $x$ cm.
Let $V$ be the corresponding volume of the box, then
$V = x(24 - 2x)(45 - 2x)$
$= x(4{x^2} - 138x + 1080)$
$= 4{x^3} - 138{x^2} + 1080x$
$\Rightarrow \cfrac{{dV}}{{dx}} = 12{x^2} - 276x + 1080$
To find the maximum or minimum volume,
$\cfrac{{dV}}{{dx}} = 0 \Rightarrow 12{x^2} - 276x + 1080 = 0$
$\Rightarrow {x^2} - 23x + 90 = 0 \Rightarrow (x - 18)(x - 5) = 0$
$\Rightarrow x = 5,x \ne 18$
$\cfrac{{{d^2}V}}{{d{x^2}}} = 24x - 276\,\,{\rm{and}}\,\,{\left( {\cfrac{{{d^2}V}}{{d{x^2}}}} \right)_{x = 5}} = 24 \times 5 - 276 < 0$
Therefore, $V$ has maximum value at $x = 5$.
So, a square of side $5{\rm{ cm}}$ should be cut from each corner for the box to have a maximum volume.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Application of Derivatives. Curated by Sachin Sharma. Free for all students.