$\begin{array}{1 1} \large\frac{-28\pi}{4+\pi} &\large\frac{112}{4+\pi} \\ \large\frac{28\pi}{4-\pi} &\large\frac{112}{4-\pi} \\ \large\frac{28\pi}{4+\pi} &\large\frac{112}{4+\pi} \\ \large\frac{28\pi}{4+\pi} &\large\frac{-112}{4+\pi} \end{array} $

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- Area of a circle=$\pi r^2$
- Perimeter=$2\pi r$

Step 1:

Let one part be of length $x$,then the other part =$28-x$

Let the part of the length $x$ be convered into a circle of radius.

$2\pi r=x$

$r=\large\frac{x}{2\pi}$

Area of the circle $=\pi r^2$

$\qquad\qquad\qquad=\pi\big(\large\frac {x}{2\pi}\big)^2$

$\qquad\qquad\qquad=\pi.\large\frac{x^2}{4\pi^2}$

$\qquad\qquad\qquad=\large\frac{x^2}{4\pi}$

Step 2:

Now second part of length $28-x$ is converted into square.

Side of square=$\large\frac{28-x}{4}$

Area of square $=\big[\large\frac{28-x}{4}\big]^2$

Total area $A=\large\frac{x^2}{4\pi}+\big(\frac{28-x}{4}\big)^2$

$\large\frac{dA}{dx}=\frac{2x}{4\pi}+\frac{2}{16}$$(28-x)(-1)$

$\qquad=\large\frac{x}{2\pi}-\frac{28-x}{8}$

Step 3:

$\large\frac{dA}{dx}$$=0$

$\Rightarrow \large\frac{x}{2\pi}-\frac{28-x}{8}$$=0$----(1)

$4x=28\pi-\pi x$

$4x=28\pi-\pi x$

$4x+\pi x=28\pi$

$x[4+\pi]=28\pi$

$x=\large\frac{28\pi}{4+\pi}$

Step 4:

Other part=$28-x=28-\large\frac{28\pi}{4+\pi}$

$\qquad\qquad\qquad\quad=\large\frac{112+28\pi-28\pi}{4+\pi}$

$\qquad\qquad\qquad\quad=\large\frac{112}{4+\pi}$

Step 5:

Differentiating (1)

$\large\frac{d^2A}{dx^2}=\frac{1}{2\pi}+\frac{1}{8}$=+ve

$A$ is minimum

When $x=\large\frac{28\pi}{4+\pi}$ & $28-x=\large\frac{112}{4+\pi}$

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