Find the radius of the base and the height of a cylinder with a volume of 27pi cm ^ 3

Find the radius of the base and the height of a cylinder with a volume of 27pi cm ^ 3, which has the smallest total surface area.

Let us assume that a cylinder with base radius R and height H has the smallest total surface. As you know, to calculate the volume (V) and area (S) of the full surface of the cylinder, there are formulas: V = π * R² * H and S = 2 * π * R * (R + H).
Using the condition of setting V = 27 * π cm³, we have: π * R² * H = 27 * π, whence H = 27 / R². Hereinafter, we will omit the units of measurement cm, cm² and cm³.
Substituting the found expression into the formula for the total surface, we get: S = S (R) = 2 * π * R * (R + 27 / R²) = 2 * π * R² + 54 * π * R-1. To fulfill the requirement of the task, we apply the differential calculus. Let’s find the first derivative of the function S (R) = 2 * π * R² + 54 * π * R-1. We have: S ‘(R) = (2 * π * R² + 54 * π * R-1)’ = 4 * π * R – 54 * π * R-2. Equating this expression to zero, we get the following equation 4 * π * R – 54 * π * R-2 = 0 or 2 * π * (R³ – 27) = 0. This equation for R has one root R = 3.
Let us investigate the sign of the derivative S ‘(R) in the vicinity of the point R = 3. To the left of the point R = 3, the inequality S’ (R) <0 is true (for example, for R = 2 the derivative S ‘(2) = 4 * π * 2 – 54 * π / 4 = (16 – 27) * (π / 2) = -11 * (π / 2) <0). Similarly, to the right of the point R = 3, the inequality S ‘(R)> 0 is true (for example, for R = 4 the derivative S’ (4) = 4 * π * 4 – 54 * π / 16 = (128 – 27) * (π / 8) = 101 * (π / 4)> 0). Thus, when passing through the point R = 3, the derivative changes sign from “-” to “+”. This means that the function S (R) at R = 3 takes on its minimum value, which is S (3) = 2 * π * 3² + 54 * π * 3-1 = 18 * π + 18 * π = 36 * π.
Thus, if the radius of the base of the cylinder is 3 cm, and its height is H = (27 / R²) cm = (27/9) cm = 3 cm.
Answer: Base radius = 3 cm and height = 3 cm.