The sides of an equilateral triangle are increasing at the rate of 2cm/sec.The rate at which the area increases ,when side is 10cm is

Question

$\begin{array}{1 1} (A)\;10cm^2/s \\ (B)\;\sqrt 3cm^2/s \\ (C)\;10\sqrt 3cm^2/s \\ (D)\;\frac{10}{3}cm^2/s \end{array} $

Answer 1

Toolbox:

• To obtain the absolute maxima or minima for the function $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.

Step 1

Given the sides the equilateral triangle = $a$ cm

Area of the equilateral triangle is

$ A = \large\frac{\sqrt 3}{4}a^2$

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

$ \large\frac{dA}{d}=\large\frac{\sqrt 3}{4}2a\large\frac{da}{dt}$

But it is given by $ \large\frac{da}{dt}=2 $ cm/sec

when $ a = 10$ cm

$ \therefore \large\frac{dA}{dt_{(a=10)}}=\large\frac{\sqrt 3}{4} \times 2 \times 10 \times 2$

$ = 10\sqrt 3\: cm^2/sec$

Hence option C is the correct answer.