Q107 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

The smaller side of the rectangle with the largest area, that can be inscribed inside a semi-circle of radius $2$ units is of length

Solution

Let length of the rectangle be $2 a$ .

And let the breadth be $b$ .

So, area $A = 2 a b$ .

Now, we have, $a^{2} + b^{2} = 1$ .

$\Rightarrow A = 2 b \sqrt{1 - b^{2}}$

Differentiating w.r.t. $b$ we get,

$\frac{d A}{d b} = 2 \sqrt{1 - b^{2}} + \frac{2 b (- 2 b)}{2 \sqrt{1 - b^{2}}}$

Putting, $\frac{d A}{d b} = 0 \Rightarrow b^{2} = \frac{1}{2} \Rightarrow b = \frac{1}{\sqrt{2}}$ .

Hence, the correct answer is option: 'D'.