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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.

1 Answer

  • 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}$$=2\pi x+2\big[\large\frac{k-2\pi x}{4}\big]\big[\large\frac{-2\pi}{4}\big]$
$\quad\;=2\pi x-\large\frac{\pi}{4}[k-2\pi x]$
$\large\frac{dA}{dx}$$=0$ at
$x=\large\frac{k\pi}{4}$$\times \large\frac{2}{4\pi+\pi^2}$
On double differentiation we get
$\large\frac{d^2A}{dx^2}=\bigg[2\pi+\large\frac{\pi^2}{2}\bigg]\Rightarrow +ve$
$A$ is least when $x=\large\frac{k}{2(\pi+4)}$
Step 4:
From (3)
Step 5:
$\quad=\large\frac{k}{\pi+4}\times \frac{2(\pi+4)}{k}$
Therefore $y = 2x \rightarrow $ proves that the sum of their areas is least when the side of square (y) = diameter, i.e, 2 times the radius (x)$
answered Aug 12, 2013 by sreemathi.v
edited Mar 20, 2014 by balaji.thirumalai

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