Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimetres, find the dimensions of the can which has the minimum surface area?
Radius = (50/π)^(1/3) cm, Height = 2(50/π)^(1/3) cm
Explanation
The problem asks for the dimensions of a closed cylindrical can with a fixed volume that minimizes the surface area. This is an optimization problem involving derivatives, similar to the methodology shown in Example 26 of the provided context. The context demonstrates finding maxima/minima by setting the first derivative to zero and verifying with the second derivative. Although the explicit formulas for volume and surface area are not defined in the text chunks, they are prerequisite knowledge for the question. The solution follows the standard calculus procedure: define variables, establish the constraint (Volume = 100), formulate the objective function (Surface Area), differentiate to find the critical point, and verify the minimum. The resulting dimensions show that the height is equal to the diameter, which aligns with the principle mentioned in Question 20 of the context regarding cylinders with optimal dimensions.
Solution Steps
Step 1: Let r be the radius and h be the height. Volume V = πr²h = 100, so h = 100/(πr²).
Step 2: Surface area S = 2πr² + 2πrh. Substitute h to get S(r) = 2πr² + 200/r.
Step 3: Differentiate S(r): S'(r) = 4πr - 200/r².
Step 4: Set S'(r) = 0 for critical points: 4πr = 200/r² ⇒ r³ = 50/π ⇒ r = (50/π)^(1/3).
Step 5: Verify minimum using second derivative: S''(r) = 4π + 400/r³. Since r > 0, S''(r) > 0, confirming minima.
Step 6: Calculate height: h = 100/(π(50/π)^(2/3)) = 2(50/π)^(1/3). Thus, h = 2r.