Let sqrt{3} be the radius and frac{pi}{3} be | vedclass

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

Question

(D) Let the radius of the cone be $R = \sqrt{3}$ and the semi-vertical angle be $\alpha = \frac{\pi}{3}$.The height of the cone is $H = R \cot(\alpha)

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the radius of the cone be $R = \sqrt{3}$ and the semi-vertical angle be $\alpha = \frac{\pi}{3}$.The height of the cone is $H = R \cot(\alpha) = \sqrt{3} \cot(\frac{\pi}{3}) = \sqrt{3} 	imes \frac{1}{\sqrt{3}} = 1$.Let the radius of the inscribed cylinder be $r$ and its height be $h$.By similar triangles, $\frac{R-r}{h} = \frac{R}{H} = \frac{\sqrt{3}}{1} = \sqrt{3}$.Thus, $h = \frac{R-r}{\sqrt{3}} = \frac{\sqrt{3}-r}{\sqrt{3}} = 1 - \frac{r}{\sqrt{3}}$.The volume of the cylinder is $V = \pi r^2 h = \pi r^2 (1 - \frac{r}{\sqrt{3}}) = \pi (r^2 - \frac{r^3}{\sqrt{3}})$.To maximize $V$, we find $\frac{dV}{dr} = \pi (2r - \frac{3r^2}{\sqrt{3}}) = \pi (2r - \sqrt{3}r^2) = 0$.Since $r 
eq 0$, we have $2 - \sqrt{3}r = 0$, so $r = \frac{2}{\sqrt{3}}$.The height of the cylinder is $h = 1 - \frac{1}{\sqrt{3}} (\frac{2}{\sqrt{3}}) = 1 - \frac{2}{3} = \frac{1}{3}$.

Let $\sqrt{3}$ be the radius and $\frac{\pi}{3}$ be the semi-vertical angle of a given cone. Then the height of the right circular cylinder of maximum volume that can be inscribed in the given cone is

Similar Questions

If $f(x) = x + \frac{1}{x}$ for $x > 0$,then its greatest value is:

The sum of the absolute maximum and absolute minimum values of the function $f(x) = \tan^{-1}(\sin x - \cos x)$ in the interval $[0, \pi]$ is.

Divide $20$ into two parts such that the product of one part and the cube of the other is maximum. The two parts are

The volume of the largest possible right circular cylinder that can be inscribed in a sphere of radius $R = \sqrt{3}$ is

$x$ and $y$ are two positive integers such that $2x + 3y = 50$. If $x^2 y^3$ is maximum for $x = \alpha$ and $y = \beta$, then $\frac{\alpha}{2} + \frac{\beta}{5} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module