The height of a right circular cylinder of ma | vedclass

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

Question

(C) Let the radius of the cylinder be $r$ and its height be $h$. The sphere has a radius $a$ (since diameter is $2a$).In the right-angled triangle for

Vedclass · Accepted Answer

$\frac{2 a}{\sqrt{3}}$

Vedclass · Answer

(C) Let the radius of the cylinder be $r$ and its height be $h$. The sphere has a radius $a$ (since diameter is $2a$).In the right-angled triangle formed by the radius of the sphere,the radius of the cylinder,and half the height of the cylinder,we have by Pythagoras theorem:$r^2 + (h/2)^2 = a^2$$r^2 = a^2 - \frac{h^2}{4}$The volume $V$ of the cylinder is given by $V = \pi r^2 h$.Substituting $r^2$ in the volume formula:$V = \pi (a^2 - \frac{h^2}{4}) h = \pi (a^2 h - \frac{h^3}{4})$To find the maximum volume,we differentiate $V$ with respect to $h$:$\frac{dV}{dh} = \pi (a^2 - \frac{3h^2}{4})$Setting $\frac{dV}{dh} = 0$ for critical points:$a^2 - \frac{3h^2}{4} = 0$$3h^2 = 4a^2$$h^2 = \frac{4a^2}{3}$$h = \frac{2a}{\sqrt{3}}$To verify it is a maximum,we check the second derivative:$\frac{d^2V}{dh^2} = \pi (0 - \frac{6h}{4}) = -\frac{3\pi h}{2}  0$.Since the second derivative is negative,the volume is maximum at $h = \frac{2a}{\sqrt{3}}$.

The height of a right circular cylinder of maximum volume inscribed in a sphere of diameter $2 a$ is

Similar Questions

What is the maximum value of $f(x) = x^3 - 18x^2 + 96x$ in the interval $(0, 9)$?

The condition for the function $f(x) = x^3 + px^2 + qx + r$ $(x \in R)$ to have no extreme value is:

If $x + y = 8$,find the maximum value of $xy$.

If the function $f(x) = (\frac{1}{x})^{2x}$ for $x > 0$ attains the maximum value at $x = \frac{1}{e}$,then:

$A$ right solid circular cylinder of given volume will have the least total surface area when

Vedclass Test Series

Exam Paper Generator

Online Exam Module