The maximum volume of the right circular cone | vedclass

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

Question

(B) $\because$ Slant height of the cone,$L = 6$ units. Let the radius be $r$ and height be $h$. Volume $(V) = \frac{1}{3} \pi r^2 h$. Since $L^2 = r^2

Vedclass · Accepted Answer

$16 \sqrt{3} \pi$ cu units

Vedclass · Answer

(B) $\because$ Slant height of the cone,$L = 6$ units. Let the radius be $r$ and height be $h$. Volume $(V) = \frac{1}{3} \pi r^2 h$. Since $L^2 = r^2 + h^2$,we have $r^2 = L^2 - h^2 = 36 - h^2$. Substituting this into the volume formula: $V = \frac{1}{3} \pi (36 - h^2) h = \frac{1}{3} \pi (36h - h^3)$. To find the maximum volume,differentiate $V$ with respect to $h$: $\frac{dV}{dh} = \frac{1}{3} \pi (36 - 3h^2)$. Setting $\frac{dV}{dh} = 0$,we get $36 - 3h^2 = 0 \Rightarrow h^2 = 12 \Rightarrow h = 2 \sqrt{3}$ units. Checking the second derivative: $\frac{d^2 V}{dh^2} = \frac{1}{3} \pi (-6h) = -2 \pi h$. At $h = 2 \sqrt{3}$,$\frac{d^2 V}{dh^2} = -4 \sqrt{3} \pi Maximum volume $V = \frac{1}{3} \pi (36 - (2 \sqrt{3})^2) (2 \sqrt{3}) = \frac{1}{3} \pi (36 - 12) (2 \sqrt{3}) = \frac{1}{3} \pi (24) (2 \sqrt{3}) = 16 \sqrt{3} \pi$ cu units.

The maximum volume of the right circular cone with slant height $6$ units is

Similar Questions

Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x-2) + \alpha x^2 + 4x - \alpha, x > 2$, is

In the interval $[-2, 4]$,the absolute maximum of $f(x) = 2x^3 - 3x^2 - 12x + 5$ occurs at $x =$

The least value of the sum of any positive real number and its reciprocal is

The maximum area (in sq. units) of a rectangle having its base on the $x-$axis and its other two vertices on the parabola $y = 12 - x^2$ such that the rectangle lies inside the parabola,is

The range of $a \in R$ for which the function $f(x)=(4 a-3)\left(x+\log _{e} 5\right)+2(a-7) \cot \left(\frac{x}{2}\right) \sin ^{2}\left(\frac{x}{2}\right)$ where $x \neq 2 n \pi, n \in N$,has critical points,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module