Question 73 of 82advanced🔧 ApplyLong Answer6 marks

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

Correct Answer

Let the radius of the circle be rr and the side of the square be xx.

Given: Sum of perimeters = kk (constant)

Perimeter of circle = 2πr2\pi r

Perimeter of square = 4x4x

Therefore,

2πr+4x=k... (1)2\pi r + 4x = k \quad \text{... (1)}

Formula: Sum of areas:

A=πr2+x2A = \pi r^2 + x^2

From equation (1):

r=k−4x2πr = \frac{k - 4x}{2\pi}

Substitution: Substituting in AA:

A=π(k−4x2π)2+x2=(k−4x)24π+x2A = \pi \left(\frac{k - 4x}{2\pi}\right)^2 + x^2 = \frac{(k - 4x)^2}{4\pi} + x^2

Calculation: Differentiating with respect to xx:

dAdx=2(k−4x)(−4)4π+2x=−2(k−4x)π+2x\frac{dA}{dx} = \frac{2(k - 4x)(-4)}{4\pi} + 2x = \frac{-2(k - 4x)}{\pi} + 2x

For minimum, dAdx=0\frac{dA}{dx} = 0:

−2(k−4x)π+2x=0\frac{-2(k - 4x)}{\pi} + 2x = 0 k−4x=πxk - 4x = \pi x k=x(4+π)k = x(4 + \pi)

Therefore,

x=k4+πx = \frac{k}{4 + \pi}

From equation (1):

r=k−4x2π=k−4k4+π2π=k2(4+π)r = \frac{k - 4x}{2\pi} = \frac{k - \frac{4k}{4 + \pi}}{2\pi} = \frac{k}{2(4 + \pi)}

Hence,

xr=k4+πk2(4+π)=2\frac{x}{r} = \frac{\frac{k}{4 + \pi}}{\frac{k}{2(4 + \pi)}} = 2

Therefore, side of square = 2 × radius of circle.

Second derivative test:

d2Adx2=8π+2>0\frac{d^2A}{dx^2} = \frac{8}{\pi} + 2 > 0

Since d2Adx2>0\frac{d^2A}{dx^2} > 0, the sum of areas is minimum when side of square is double the radius of circle.

∴ Hence proved.

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 7 | (Page 38)
For More Understanding

Explanation

This is a standard optimization problem from NCERT Class 12 Mathematics. The solution follows the textbook approach: express one variable in terms of the other using the constraint equation, substitute into the objective function (sum of areas), differentiate, find critical points, and verify using the second derivative test.

The key insight is that when the derivative equals zero, we get x=2rx = 2r, and the positive second derivative confirms it's a minimum.

Solution Steps

  1. Step 1: Define variables - Let radius of circle = r, side of square = x

  2. Step 2: Write constraint equation - 2πr + 4x = k

  3. Step 3: Express sum of areas A = πr² + x²

  4. Step 4: Substitute r = (k - 4x)/(2π) into A

  5. Step 5: Differentiate A with respect to x and set dA/dx = 0

  6. Step 6: Solve to get x = k/(4 + π) and r = k/(2(4 + π))

  7. Step 7: Show x/r = 2, proving side = 2 × radius

  8. Step 8: Verify minimum using second derivative test (d²A/dx² > 0)

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