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(B) Let the height of the solid cylinder be $h$ and its radius be $r$.The volume $V$ is given by $V = \pi r^2 h$, which implies $h = \frac{V}{\pi r^2}

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its height is equal to its diameter.

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(B) Let the height of the solid cylinder be $h$ and its radius be $r$.The volume $V$ is given by $V = \pi r^2 h$, which implies $h = \frac{V}{\pi r^2}$.The total surface area $S$ is given by $S = 2\pi rh + 2\pi r^2$.Substituting $h$ in terms of $V$ and $r$: $S = 2\pi r \left(\frac{V}{\pi r^2}\right) + 2\pi r^2 = \frac{2V}{r} + 2\pi r^2$.To find the minimum surface area, differentiate $S$ with respect to $r$: $\frac{dS}{dr} = -\frac{2V}{r^2} + 4\pi r$.Setting $\frac{dS}{dr} = 0$, we get $4\pi r = \frac{2V}{r^2}$, which implies $V = 2\pi r^3$.Substituting $V = \pi r^2 h$ into this equation: $\pi r^2 h = 2\pi r^3$, which simplifies to $h = 2r$.Since $2r$ is the diameter, the surface area is least when the height is equal to the diameter.

$A$ right solid circular cylinder of given volume will have the least total surface area when

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