Applications of Derivatives — Maths STD 12 Science — Question

Let x cm and y cm be the length and breadth of the rectangle.Then its area is xy = 50
$y=\frac{50}{x}$
Perimeter of the rectangle= 2(x + y)
$=2\left(x+\frac{50}{x}\right)$
$\operatorname{Let} f(x)=2\left(x+\frac{50}{x}\right)$
then $f^{\prime}(x)=2\left(1-\frac{50}{x^2}\right)$ and $f^{\prime \prime}(x)=2\left(\frac{100}{x^3}\right)$
$f^{\prime}(x)=0$, then $2\left(1-\frac{50}{x^2}\right)=0$
$\begin{aligned} & x^2=50 \\ & x= \pm 5 \sqrt{2}\end{aligned}$
But $x$ can not be negative and hence $x=5 \sqrt{2}$
and $f^{\prime \prime}(5 \sqrt{2})=\frac{200}{(5 \sqrt{2})^3}>0$
$f$ has a minimum value at $x=5 \sqrt{2}$
$F$ or $x=5 \sqrt{2}, y=\frac{50}{5 \sqrt{2}}=5 \sqrt{2}$
$x=5 \sqrt{2} cm , y=5 \sqrt{2} cm$