Question

The equal sides of an isosceles triangle are each equal to $2011cm$. The length of the third side (in $cm$) which makes the area of the triangle maximum is

• A. $2011$
• B. $\sqrt{2}\times 2011$
• C. $\sqrt{3}\times 2011$
• D. $4022$

Answer 1

The correct option is B $\sqrt{2}\times 2011$

We have,

$\begin{array}{c}l=\sqrt{a^2-\frac{b^2}{4}}\\ A=\frac{1}{2}\left(b\times l\right)\\ \frac{1}{2}b\sqrt{a^2-\frac{b^2}{4}}\end{array}$

For max/ min $\frac{dA}{dB}=\frac{1}{2}$

$\begin{array}{c}l=\sqrt{a^2-\frac{b^2}{4}}\\ A=\frac{1}{2}\left(b\times l\right)\\ \frac{1}{2}b\sqrt{a^2-\frac{b^2}{4}}\\ \frac{dA}{dB}=\frac{1}{2}\\ A^2=\frac{1}{4}b\sqrt{a^2-\frac{b^2}{4}}=\frac{1}{4}{a}^2{b}^2-\frac{b^4}{16}\\ 2A\frac{dA}{db}=\frac{1}{4}{a}^2\times 2b.\frac{b^3}{4}\\ a^2\times 2b=b^3.b^2\\ \frac{\sqrt{2a}}{b}=b\\ b=\sqrt{2}\times 2011\end{array}$

$\therefore$ Option $B$ is correct.