The height (in cm) of a cylinder of the great | vedclass

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

Question

(B) Let $R$ be the radius of the sphere and $r$ and $h$ be the radius and height of the cylinder, respectively.From the geometry of the sphere and ins

Vedclass · Accepted Answer

$2 \sqrt{3}$

Vedclass · Answer

(B) Let $R$ be the radius of the sphere and $r$ and $h$ be the radius and height of the cylinder, respectively.From the geometry of the sphere and inscribed cylinder, we have the relation: $R^2 = r^2 + (h/2)^2$.Thus, $r^2 = R^2 - h^2/4$.The volume $V$ of the cylinder is given by $V = \pi r^2 h = \pi (R^2 - h^2/4) h = \pi R^2 h - \pi h^3/4$.To find the maximum volume, we differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4}$.Setting $\frac{dV}{dh} = 0$, we get $\pi R^2 = \frac{3\pi h^2}{4}$, which implies $h^2 = \frac{4R^2}{3}$, so $h = \frac{2R}{\sqrt{3}}$.Given $R = 3 \ cm$, we have $h = \frac{2 	imes 3}{\sqrt{3}} = 2 \sqrt{3} \ cm$.Since $\frac{d^2V}{dh^2} = -\frac{6\pi h}{4}  0$, the volume is maximum at $h = 2 \sqrt{3} \ cm$.

The height (in $cm$) of a cylinder of the greatest volume that can be inscribed in a sphere of radius $3 \ cm$ is

Similar Questions

If $x$ and $y$ are two positive integers such that $x+y=24$ and $x^3 y^5$ is maximum, then $x^2+y^2=$

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

If $f(x) = 7e^{\sin^2 x} - e^{\cos^2 x} + 2$,then $\sqrt{7f_{\min} + f_{\max}}$ is equal to

The maximum area of a rectangle inscribed in a circle of radius $10 \text{ cm}$ is

If $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has local maximum and minimum at $x = p$ and $x = q$,respectively,then $(p, q)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module