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)

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)
$A=\sqrt{9\left(9-x\right)\left(9-x\right)\left[9-\left(18-2x\right)\right]}$,
$\frac{9}{2}<x<9$ (Heron's method)
$\Rightarrow A=3\left(9-x\right)\sqrt{2x-9}$,
$x\in \left(\frac{9}{2},9\right)$
$\therefore \frac{dA}{dx}=3\left(\left(9-x\right)\cdot \frac{1}{2\sqrt{2x-9}}\cdot 2+\sqrt{2x-9}\left(-1\right)\right)$
$=3\left(\frac{9-x-2x+9}{\sqrt{2x-9}}\right)$
$=\frac{9\left(6-x\right)}{\sqrt{2x-9}}$, $x\in \left(\frac{9}{2},9\right)$
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 $=18-2\times 6=6$ i.e. the triangle is equilateral)