# The sum of the perimeter of a circle and square is $$k$$, where $$k$$ is some constant. Prove that the sum of their areas is least when the side of square is double the radius of the circle.

Toolbox:
• Circumference of the circle$=2\pi r$
• Perimeter of the square$=4s$
• Area of the circle$=\pi r^2$
• Area of the Square$=s^2$
Step 1:
Let $x=$Radius of the circle.
$y$=Side of the square.
Circumference of the circle$=2\pi r=2\pi x$
Perimeter of square$=4s=4y$
Sum of perimeters of circle and square=$2\pi x+4y=k$----(1)
Area of circle $=\pi r^2=\pi x^2$
Area of square$=y^2(i.e\; s^2)$
Step 2:
Sum of areas of circle & square=$\pi x^2+y^2$
$A=\pi x^2+y^2$----(2)
From (1)$2\pi x+4y=k$
$4y=k-2\pi x$
$y=\large\frac{k-2\pi x}{4}$-----(3)
$A=\pi x^2+\bigg[\large\frac{k-2\pi x}{4}\bigg]^2$[By substituting the value of y]
Step 3:
Differentiating with respect to $x$ we get,
$\large\frac{dA}{dx}$$=2\pi x+2\big[\large\frac{k-2\pi x}{4}\big]\big[\large\frac{-2\pi}{4}\big] \quad\;=2\pi x-\large\frac{\pi}{4}[k-2\pi x] \quad\;=[2\pi+\large\frac{\pi^2}{2}]x-\large\frac{k\pi}{4} \large\frac{dA}{dx}$$=0$ at