IIT JEE 1983 Maths - MCQ Question 2

As we know angle in a semicircle is a right angle . so let the sides be a, b and c

Quite obviously a^2+b^2 = c^2 wherein c = diameter of circle

Now area = 1/2 * a*b

By AM GM It obviously follows that

c^2/2 is greater than or equal to ab

Hence area is less than or equal to c^2/4 and equality holds when triangle is isosceles.

Therefore Triangle has maximal area when the triangle is isosceles

If AB is a diameter and C is any point on the circumference, then $\bigtriangleup ABC$ is a right triangle. Let $AB = L, AC = x, BC = \sqrt{L^2 - x^2}.$ , and define the following functions for area, $A(x),$ , and perimeter, $P(x):$

$A(x) = \frac{1}{2}x \sqrt{L^2 - x^2}, P(x) = L + x + \sqrt{L^2 - x^2}$

and taking the first derivatives of each gives:

$\frac{dA}{dx} = \frac{1}{2} [\sqrt{L^2 - x^2} - \frac{x^2}{\sqrt{L^2 - x^2}}]$

$\frac{dP}{dx} = 1 - \frac{x}{\sqrt{L^2 - x^2}}$

which setting each expression equal to zero yields: $\frac{dA}{dx} = 0 \Rightarrow x = \frac{L}{\sqrt{2}}; \frac{dP}{dx} = 0 \Rightarrow x = \frac{L}{\sqrt{2}}.$

hence, the critical triangle is right-isosceles. Taking the second derivatives of each (evaluated at the common critical point $x = \frac{L}{\sqrt{2}}$ ) yields:

$\frac{d^2A}{dx^2} = \frac{-3x}{2 \cdot \sqrt{L^2 - x^2}} \Rightarrow A''(\frac{L}{\sqrt{2}}) < 0$

$\frac{d^2P}{dx^2} = \frac{-L^2}{(L^2 - x^2)^{\frac{3}{2}}} \Rightarrow P''(\frac{L}{\sqrt{2}}) < 0.$

hence, $\bigtriangleup ABC$ has both a maximum area and a maximum perimeter when it is right-isosceles. This makes choice D correct.