The height of a right circular cone of maximu | vedclass

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

Question

(A) Let $R = a/2$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.From the geometry of th

Vedclass · Accepted Answer

$(2/3)a$

Vedclass · Answer

(A) Let $R = a/2$ be the radius of the sphere. Let $h$ be the height of the cone and $r$ be the radius of the base of the cone.From the geometry of the sphere,if the center of the sphere is at a distance $x$ from the base of the cone,then $h = R + x = a/2 + x$.The radius of the cone base is $r^2 = R^2 - x^2 = (a/2)^2 - x^2$.The volume of the cone is $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi ((a/2)^2 - x^2)(a/2 + x)$.$V(x) = \frac{1}{3} \pi (a/2 - x)(a/2 + x)(a/2 + x) = \frac{1}{3} \pi (a/2 - x)(a/2 + x)^2$.To find the maximum volume,differentiate $V$ with respect to $x$ and set to $0$:$\frac{dV}{dx} = \frac{1}{3} \pi [(-1)(a/2 + x)^2 + (a/2 - x) \cdot 2(a/2 + x)] = 0$.$(a/2 + x) [-(a/2 + x) + 2(a/2 - x)] = 0$.Since $h 
eq 0$,$a/2 + x 
eq 0$,so $-(a/2 + x) + a - 2x = 0$.$-a/2 - x + a - 2x = 0 \implies a/2 = 3x \implies x = a/6$.The height $h = a/2 + x = a/2 + a/6 = 3a/6 + a/6 = 4a/6 = (2/3)a$.

The height of a right circular cone of maximum volume inscribed in a sphere of diameter $a$ is-

Similar Questions

Let $f: R \rightarrow R$ be given by $f(x) = |x^2 - 1|$, then

If the function $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$,for some $a \in R$,is increasing in $(0, 1]$ and decreasing in $[1, 5)$,then a root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ $(x \neq 1)$ is

The maximum value of $\frac{\log x}{x}, 0 < x < \infty$ is

The radius of the cylinder of maximum volume,which can be inscribed in a sphere of radius $R$ is :-

Two particles move in the same straight line starting at the same moment from the same point in the same direction. The first moves with constant velocity $u$ and the second starts from rest with constant acceleration $f$. Then,

Vedclass Test Series

Exam Paper Generator

Online Exam Module