Let the radius and height of a right circular | vedclass

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(A) The volume of a right circular cylinder is given by $V = \pi r^2 h$. Given the relation $r^2 + h = 6$,we can express $h$ as $h = 6 - r^2$. Substit

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$\frac{1}{\sqrt{3}}$

Vedclass · Answer

(A) The volume of a right circular cylinder is given by $V = \pi r^2 h$. Given the relation $r^2 + h = 6$,we can express $h$ as $h = 6 - r^2$. Substituting this into the volume formula: $V = \pi r^2 (6 - r^2) = \pi (6r^2 - r^4)$. To find the maximum volume,we differentiate $V$ with respect to $r$: $\frac{dV}{dr} = \pi (12r - 4r^3)$. Setting $\frac{dV}{dr} = 0$,we get $12r - 4r^3 = 0$,which implies $4r(3 - r^2) = 0$. Since $r > 0$,we have $r^2 = 3$,so $r = \sqrt{3}$. Substituting $r^2 = 3$ into the relation $h = 6 - r^2$,we get $h = 6 - 3 = 3$. Therefore,the ratio $\frac{r}{h} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$.

Let the radius and height of a right circular cylinder be related as $r^2 + h = 6$. If the volume of the cylinder is maximum,then the value of $\frac{r}{h}$ is:

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