# 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

**Solution:**

Let r be the radius of the circle and a be the side of the square

Then, we have:

2π r + 4a = k

a = (k - 2πr) / 4

The sum of the areas of the circle and the square (A) is given by,

A = π r^{2} + a^{2}

= πr^{2} + (k - 2πr)^{2 }/ 16

Hence,

dA/dr = 2π r + (2 (k - 2πr)(- 2π)) / 16

= 2πr - π(k - 2πr)/4

Now,

dA/dr = 0

⇒ 2πr - π(k - 2π r) / 4 = 0

2πr = π (k - 2π r) / 4

⇒ 8r = k - 2πr

⇒ (8 + 2π) r = k

r = k/(8 + 2π)

r = k/2(4 + π)

When, r = k / 2(4 + π),

⇒ d^{2}A/dr^{2} > 0

The sum of the areas is least when,

r = k / 2(4 + π)

When, r = k / 2(4 + π)

Then,

a = k - 2π [k / (4 + π)] / 4

= (k (4 + π) - πk) / 4 (4 + π)

= 4k / 4 (4 + π)

= 2r

Hence, it has been proved that the sum of their areas is least when the side of the square is double the radius of the circle

NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 10

## 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.

**Summary:**

Given that the sum of the perimeter of a circle and square is k, where k is some constant. Hence we have proved that the sum of their areas is least when the side of the square is double the radius of the circle

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