Maximaand Minima

4 Marks Questions

1. The sides of an equilateral triangle are increasing at the rate of 2cm/s. Find the rate at which the area increases, when the side is

10 cm. All India 2014C

Let the side of triangle be a.

<math>\frac{da}{dt} = 2 \text{ cm/s}</math>

[given] (1)

Now, area of equilateral triangle having side a is given by

<math display="block">A = \frac{\sqrt{3}a^2}{4}</math> (1) On differentiating w.r.t.t, we get

<math display="block">\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot (2a) \frac{da}{dt}</math> (1)

On putting <math>\frac{da}{dt} = 2</math> cm/s and <math>a = 10</math> cm, we get

<math>\frac{dA}{dt} = \frac{\sqrt{3}}{4} \times 2 \times 10 \times 2 = 10\sqrt{3} \text{ cm}^2/\text{s}</math> (1)

1. The sum of the perimeters 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 the square is double the radius of the

circle. Delhi 2014C; All India 2008

Let r be the radius of circle and x be the side of a square. Then, given that

Perimeter of square + Circumference of circle (1)

i.e. <math>4x + 2\pi r = k</math>

...(i) (1)

<math>x = \frac{k - 2\pi r}{4}</math>

Let A denotes the sum of their areas. :: <math display="block">A = x^2 + \pi r^2 \qquad \dots (ii)</math>

<math display="block">\begin{bmatrix} \because \text{ area of a square} = (\text{Side})^2 \\ \text{and area of circle} = \pi r^2 \end{bmatrix}</math>

On putting the value of x from Eq. (i) in Eq.(ii), we get

(1 - \2 <math display="block">A = \left(\frac{k - 2\pi r}{4}\right)^2 + \pi r^2</math>

On differentiating w.r.t. r, we get

<math display="block">\frac{dA}{dr} = 2\left(\frac{k-2\pi r}{4}\right)\left(-\frac{2\pi}{4}\right) + 2\pi r</math>

<math display="block">= -\frac{\pi}{4}(k-2\pi r) + 2\pi r</math> (1)

For maxima and minima, put <math>\frac{dA}{dr} = 0</math> <math>\Rightarrow</math> <math display="block">-\frac{\pi}{4}(k - 2\pi r) + 2\pi r = 0</math>