Question

Differentiate its volume with respect to radius of the base, where the surface area is constant.

Answer 1

Concept:
• Volume of cylinder: \( V = \pi r^2 h \).
• Surface area of an open cylinder: \( S = \pi r^2 + 2\pi r h \).
• To differentiate \( V \) wrt \( r \), express \( V \) as a function of \( r \) alone by eliminating \( h \).

Step 1: Express \( h \) in terms of \( S \) and \( r \)
From the surface area formula (where \( S \) is constant): \[ 2\pi r h = S - \pi r^2 \] \[ h = \frac{S - \pi r^2}{2\pi r} \]

Step 2: Substitute \( h \) into the volume formula
\[ V = \pi r^2 \left( \frac{S - \pi r^2}{2\pi r} \right) \] \[ V = \frac{r(S - \pi r^2)}{2} \] \[ V = \frac{Sr}{2} - \frac{\pi r^3}{2} \]

Step 3: Differentiate \( V \) with respect to \( r \)
Treating \( S \) and \( \pi \) as constants: \[ \frac{dV}{dr} = \frac{d}{dr} \left( \frac{Sr}{2} - \frac{\pi r^3}{2} \right) \] \[ \frac{dV}{dr} = \frac{S}{2} - \frac{3\pi r^2}{2} \]

Answer 2

Concept:
• For maximum volume, the derivative \( dV/dr \) must be zero.
• Use the result from the previous part to find the optimal dimensions.

Step 1: Set the derivative to zero
From part (i): \( \frac{dV}{dr} = \frac{S}{2} - \frac{3\pi r^2}{2} \). For maximum volume: \[ \frac{S}{2} - \frac{3\pi r^2}{2} = 0 \] \[ S = 3\pi r^2 \]

Step 2: Substitute the original formula for \( S \)
We know \( S = \pi r^2 + 2\pi r h \). Substituting this into our optimality condition: \[ \pi r^2 + 2\pi r h = 3\pi r^2 \]

Step 3: Simplify the relation
\[ 2\pi r h = 3\pi r^2 - \pi r^2 \] \[ 2\pi r h = 2\pi r^2 \] Dividing by \( 2\pi r \) (since \( r \neq 0 \)): \[ h = r \]

Step 4: Conclusion
To maximize the volume of an open cylindrical tumbler for a fixed surface area, the height must be equal to the radius of the base.