The smaller side of the rectangle with the la | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the radius of the semi-circle be $R = 2$. Let the rectangle have length $2x$ and breadth $y$ inscribed in the semi-circle.By the Pythagorean t

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) Let the radius of the semi-circle be $R = 2$. Let the rectangle have length $2x$ and breadth $y$ inscribed in the semi-circle.By the Pythagorean theorem,$x^2 + y^2 = R^2 = 2^2 = 4$.Thus,$x = \sqrt{4 - y^2}$.The area of the rectangle is $A = (2x) 	imes y = 2y \sqrt{4 - y^2}$.To maximize $A$,we maximize $A^2 = 4y^2(4 - y^2) = 16y^2 - 4y^4$.Let $f(y) = 16y^2 - 4y^4$. Differentiating with respect to $y$,we get $f'(y) = 32y - 16y^3$.Setting $f'(y) = 0$,we have $16y(2 - y^2) = 0$.Since $y > 0$,$y^2 = 2$,so $y = \sqrt{2}$.Then $x = \sqrt{4 - 2} = \sqrt{2}$.The sides of the rectangle are $2x = 2\sqrt{2}$ and $y = \sqrt{2}$.The smaller side is $\sqrt{2}$.

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

Similar Questions

The acceleration $f \text{ ft/sec}^2$ of a particle after a time $t \text{ sec}$ starting from rest is given by $f = 6 - \sqrt{1.2t}$. Then the maximum velocity $v$ and time $T$ to attain this velocity are:

For all real $x$,the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

Local maximum and local minimum values of the function $f(x) = (x - 1)(x + 2)^2$ are

The number of values of $x$ where the function $f(x) = \cos x + \cos (\sqrt{2} x)$ attains its maximum is

Prove that the function $h(x) = x^{3} + x^{2} + x + 1$ does not have any local maxima or local minima.

Vedclass Test Series

Exam Paper Generator

Online Exam Module