Question

Find the area of the largest isosceles triangle having perimeter $18$ metres.

• A. $9\sqrt{3}$
• B. $8\sqrt{3}$
• C. $4\sqrt{3}$
• D. $7\sqrt{3}$

Answer 1

Answer: (A)

$9\sqrt{3}$

Answer 2

Let $A$ be the area of the triangle when one of its equal sides is $x$ so that the base $= 18 - x - x = 18 - 2x$, $\frac{9}{2} < x < 9$ ($\because x > 0, 18 - 2x > 0$ and $x + x > 18 - 2x$ as the sum of two sides > third side) $A = \sqrt{ 9 \left(9 -x\right) \left(9 - x \right)\left[ 9 - \left( 1 8 - 2x\right)\right]}$, $\frac{9}{2} < x < 9$ (Heron's method) $\Rightarrow A = 3\left(9-x\right)\sqrt{2x - 9}$, $x\in \left(\frac{9}{2}, 9\right)$ $\therefore \frac{dA}{dx} = 3\left( \left(9-x\right)\cdot \frac{1}{2\sqrt{2x-9}}\cdot2+\sqrt{2x - 9}\left(-1\right)\right)$ $= 3\left(\frac{9-x-2x+9}{\sqrt{2x-9}} \right)$ $= \frac{9\left(6-x\right)}{\sqrt{2x-9}}$, $x\in \left(\frac{9}{2}, 9\right)$ Now, $\frac{dA}{dx} = 0$ $\Rightarrow \frac{9\left(6-x\right)}{\sqrt{2x-9}} = 0$ $\Rightarrow 6 - x = 0$ $\Rightarrow x = 6$ When $x < 6$ slightly, then $\frac{dA}{dx} = \frac{+ve\left(+ ve\right)}{+ve} = +ve$ and when $x > 6$ slightly, then $\frac{dA}{dx} = \frac{+ve\left(- ve\right)}{+ve} =- ve$ $\Rightarrow \frac{dA}{dx}$ changes sign from $+ve$ to $-ve$ as we move from shghly $< 6$ to shghtly $> 6$ through $6$ $\Rightarrow$ A has a local maximum at $x = 6$. But, $A = 3\left(9-x\right) \sqrt{2x - 9}$ is continuous in $\left(\frac{9}{2}, 9\right)$ and has only one extremum (at $x = 6$), therefore, $x = 6$ is the point of absolute maximum of $A$. Hence, $A$ is maximum when $x = 6$ and maximum value of $A = 3\left(9 - 6\right)\sqrt{2 \times 6 - 9} = 9\sqrt{3}$ (Note that when $A$ is maximum, the base of the triangle $= 1 8 - 2 \times 6 = 6$ i.e. the triangle is equilateral)