Question
Show that of all the rectangles inscribed in a given fixed circle, the square has maximum area.
✓
Answer
Let PQRS be the rectangle inscribed in a given circle with centre O and radius a.
Let x and y be the length and breadth of the rectangle, i. e., $x > 0$ and $y > 0.$
In right angled triangle $PQR,$ using Pythagoras theorem,
$PQ^2 + QR^2 = PR^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=(2\text{a)}^2\ \Rightarrow\ \text{y}^2=4\text{a}^2-\text{x}^2\ \Rightarrow$
$\text{y}=\sqrt{4\text{a}^2-\text{x}^2}\ \dots\text{(i)}$
Let A be the area of the rectangle, then $\ \text{A }=\text{xy}=\text{x}\sqrt{4\text{a}^2-\text{x}^2}$
$\Rightarrow\ \frac{\text{d}\text{A}}{\text{dx}}=\sqrt{4\text{a}^2-\text{x}^2}+\text{x}\frac{1}{2\sqrt{4\text{a}^2-\text{x}^2}}(-2\text{x)}$ $=\sqrt{4\text{a}^2-\text{x}^2}-\frac{\text{x}^2}{\sqrt{4\text{a}^2-\text{x}^2}}=\frac{4\text{a}^2-2\text{x}^2}{\sqrt{4\text{a}^2-\text{x}^2}}$
And $\frac{\text{d}^2\text{A}}{\text{dx}^2}=\frac{\sqrt{4\text{a}^2-\text{x}^2}(-4\text{x})-(4\text{a}^2-2\text{x}^2)\frac{(-2\text{x)}}{2\sqrt{4\text{a}^2-\text{x}^2}}}{(4\text{a}^2-2\text{x}^2)}$ $=\frac{(4\text{a}^2-2\text{x}^2)(-4\text{x)}+\text{x}(4\text{a}^2-2\text{x}^2)}{(4\text{a}^2-2\text{x}^2)^{\frac{3}{2}}}$
$\Rightarrow\ \ \frac{\text{d}^2\text{A}}{\text{dx}^2}=\frac{-12\text{a}^2\text{x}+2\text{x}^3}{(4\text{a}^2-2\text{x}^2)}=\frac{-2(6\text{a}^2-\text{x}^2)}{(4\text{a}^2-2\text{x}^2)^{\frac{3}{2}}}$
Now $ \frac{\text{d}\text{A}}{\text{dx}}=0\ \Rightarrow\ \frac{4\text{a}^2-2\text{x}^2}{\sqrt{4\text{a}^2-\text{x}^2}}$ $\Rightarrow\ \ 4\text{a}^2-2\text{x}^2=0\ \Rightarrow\ \text{x}=\sqrt{2\text{a}}$
$\therefore\ \text{At }\text{x}=\sqrt{2\text{a}},\ \frac{\text{d}^2\text{A}}{\text{dx}^2}=\frac{-2\big(\sqrt{2\text{a}\big)}(6\text{a}^2-2\text{a}^2)}{2\sqrt{2\text{a}^3}}$ $=\frac{-8\sqrt{2\text{a}^3}}{2\sqrt{2\text{a}^3}}=-4\ [\text{Negative}]$
$\therefore\ \text{At}\text{x}=\sqrt{2\text{a}},$ area of rectangle is maximum.
And from eq.(i) $\text{y}=\sqrt{4\text{a}^2-2\text{a}^2}=\sqrt{2\text{a}}, \text{i. e.,}\text{x}=\text{y}=\sqrt{2\text{a}}$
Therefore, the area of inscribed rectangle is maximum when it is square.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate the definite integral in Exercise:
$\int^{\pi}_{0}\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
→$\int^{\pi}_{0}\frac{\text{x}\tan\text{x}}{\sec\text{x}+\tan\text{x}}\text{dx}$
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
$f(x) = (x - 1)^2 + 3$ in $[-3, 1]$
→$f(x) = (x - 1)^2 + 3$ in $[-3, 1]$
$\text{Find}: \int(x + 3) \sqrt{3 - 4x - x^{2}} dx.$
→Find the vector and cartesian equations of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=6$
→The probability of a man hitting a target is 0.25. He shoots 7 times. What is the probability of his hitting at least twice?
A coin is tossed thrice and all the eight outcomes are assumed equally likely. In which of the following cases are the following events A and B are independent?
A = the number of heads is two,
B = the last throw results in head.
A = the number of heads is two,
B = the last throw results in head.
A box has 20 pens of which 2 are defective. Calculate the probability that out of 5 pens drawn one by one with replacement, at most 2 are defective.
In a shop X, 30 tins of pure ghee and 40 tins of adulterated ghee which look alike, are kept for sale while in shop Y, similar 50 tins of pure ghee and 60 tins of adulterated ghee are there. One tin of ghee is purchased from one of the randomly selected shops and is found to be adulterated. Find the probability that it is purchased from shop Y. What measures should be taken to stop adulteration?
Show that the relation R on the set Z of integers, given by R = {(a, b): 2 divides a - b}, is an equivalence relation.
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
→$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$