The maximum volume of a right circular cylind | vedclass

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(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 $\pi \text{ m}^3$)

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