The maximum area (in sq. units) of a rectangl | vedclass

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

Question

(C) Let the vertices of the rectangle on the parabola be $(\alpha, 12 - \alpha^2)$ and $(-\alpha, 12 - \alpha^2)$,where $\alpha > 0$.The base of the r

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Let the vertices of the rectangle on the parabola be $(\alpha, 12 - \alpha^2)$ and $(-\alpha, 12 - \alpha^2)$,where $\alpha > 0$.The base of the rectangle lies on the $x-$axis,so the length of the base is $2\alpha$ and the height is $12 - \alpha^2$.The area $A$ of the rectangle is given by $A = 	ext{length} 	imes 	ext{height} = 2\alpha(12 - \alpha^2) = 24\alpha - 2\alpha^3$.To find the maximum area,we differentiate $A$ with respect to $\alpha$:$\frac{dA}{d\alpha} = 24 - 6\alpha^2$.Setting $\frac{dA}{d\alpha} = 0$,we get $24 - 6\alpha^2 = 0$,which implies $\alpha^2 = 4$,so $\alpha = 2$ (since $\alpha > 0$).To verify this is a maximum,we check the second derivative: $\frac{d^2A}{d\alpha^2} = -12\alpha$. At $\alpha = 2$,$\frac{d^2A}{d\alpha^2} = -24 The maximum area is $A = 2(2)(12 - 2^2) = 4(12 - 4) = 4(8) = 32$ sq. units.

The maximum area (in sq. units) of a rectangle having its base on the $x-$axis and its other two vertices on the parabola $y = 12 - x^2$ such that the rectangle lies inside the parabola,is

Similar Questions

Maximum value of $x(1 - x)^2$ when $0 \le x \le 2$,is

Let $f(x)=1+\frac{x}{1 !}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\frac{x^4}{4 !}$. The number of real roots of $f(x)=0$ is

If $ax + \frac{b}{x} \ge c$ for all positive $x$,where $a, b > 0$,then:

Find all points of local maxima and local minima of the function $f$ given by $f(x) = x^3 - 3x + 3$.

Find the local maximum and local minimum values for the function $f(x) = x\sqrt{1 - x}$ where $0 < x < 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module