The lateral edge of a regular rectangular pyr | vedclass

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

Question

(C) Let the height of the pyramid be $h$ and the distance from the center of the base to a vertex be $x$. Given the lateral edge $a$,we have $h = a \s

Vedclass · Accepted Answer

$\cot^{-1}\sqrt{2}$

Vedclass · Answer

(C) Let the height of the pyramid be $h$ and the distance from the center of the base to a vertex be $x$. Given the lateral edge $a$,we have $h = a \sin \alpha$ and $x = a \cos \alpha$.The base is a square with diagonal $2x$. The area of the square base is $A = \frac{1}{2} 	imes (	ext{diagonal})^2 = \frac{1}{2} 	imes (2x)^2 = 2x^2$.The volume $V$ of the pyramid is $V = \frac{1}{3} A h = \frac{1}{3} (2x^2) h$.Substituting $x = a \cos \alpha$ and $h = a \sin \alpha$:$V(\alpha) = \frac{2}{3} (a \cos \alpha)^2 (a \sin \alpha) = \frac{2}{3} a^3 \sin \alpha \cos^2 \alpha$.To maximize $V$,we differentiate with respect to $\alpha$ and set to $0$:$V'(\alpha) = \frac{2}{3} a^3 [\cos \alpha \cdot \cos^2 \alpha + \sin \alpha \cdot 2 \cos \alpha (-\sin \alpha)] = 0$.$\cos^3 \alpha - 2 \sin^2 \alpha \cos \alpha = 0$.Since $\cos \alpha 
eq 0$,we have $\cos^2 \alpha = 2 \sin^2 \alpha$,which implies $	an^2 \alpha = \frac{1}{2}$,or $	an \alpha = \frac{1}{\sqrt{2}}$.Thus,$\alpha = 	an^{-1}(\frac{1}{\sqrt{2}}) = \cot^{-1}(\sqrt{2})$.

The lateral edge of a regular rectangular pyramid is $a \text{ cm}$ long. The lateral edge makes an angle $\alpha$ with the plane of the base. The value of $\alpha$ for which the volume of the pyramid is greatest,is

Similar Questions

The maximum volume of a right circular cylinder if the sum of its radius and height is $6 \text{ m}$ is: (in $\pi \text{ m}^3$)

The ordinates of the points on the curve $y = \tan^{-1}(\sin \sqrt{x})$,$0 \leq x \leq 8\pi^2$,at which the tangent is parallel to the $x$-axis are

If $(\alpha, \beta)$ and $(\gamma, \delta)$ where $\alpha < \gamma$ are the turning points of $f(x) = 2x^3 - 15x^2 + 36x - 8$,then $\alpha - \gamma - \beta + \delta =$

The number of real roots of the equation $e^{4x} + 2e^{3x} - e^{x} - 6 = 0$ is:

The triangle of maximum area that can be inscribed in a given circle of radius $r$ is ...... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module