The maximum volume of a right circular cylind | vedclass

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

Question

(B) Let the radius of the cylinder be $r$ and the height be $h$. Given $r + h = 6$, so $h = 6 - r$.The volume $V$ of the cylinder is given by $V = \pi

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Let the radius of the cylinder be $r$ and the height be $h$. Given $r + h = 6$, so $h = 6 - r$.The volume $V$ of the cylinder is given by $V = \pi r^2 h$.Substituting $h$, we get $V(r) = \pi r^2 (6 - r) = \pi (6r^2 - r^3)$.To find the maximum volume, we differentiate $V$ with respect to $r$ and set it to zero:$\frac{dV}{dr} = \pi (12r - 3r^2) = 0$.$3r(4 - r) = 0$, which gives $r = 0$ (not possible) or $r = 4$.For $r = 4$, $h = 6 - 4 = 2$.Checking the second derivative: $\frac{d^2V}{dr^2} = \pi (12 - 6r)$. At $r = 4$, $\frac{d^2V}{dr^2} = \pi (12 - 24) = -12\pi The maximum volume is $V = \pi (4)^2 (2) = 32\pi 	ext{ m}^3$.

The maximum volume of a right circular cylinder if the sum of its radius and height is $6 \text{ m}$ is: (in $\pi \text{ m}^3$)

Similar Questions

For all $a, b \in R$,the function $f(x) = 3x^4 - 4x^3 + 6x^2 + ax + b$ has:

If $a$ and $b$ are positive numbers such that $a > b$,then the minimum value of $a \sec \theta - b \tan \theta$ for $0 < \theta < \frac{\pi}{2}$ is

Let $f: R \rightarrow R$ be given by $f(x) = |x^{2} - 1|$, $x \in R$. Then:

On the interval $[0, 1]$, the function $f(x) = x^{25}(1 - x)^{75}$ attains its maximum value at the point $x = ....$

If $y = \frac{ax - b}{(x - 1)(x - 4)}$ has a turning point $P(2, -1)$,then the values of $a$ and $b$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module