A soft drink can has a volume of 340 cm³, a height of h cm and a radius of r cm - NSC Mathematics - Question 9 - 2016 - Paper 1

The surface area A of a cylinder is given by: $A = 2\pi r^2 + 2\pi rh$ Substituting for h using our previous result: $A = 2\pi r^2 + 2\pi r \left(\frac{340}{\pi r^2}\right)$ After simplifying, we find: $A = 2\pi r^2 + \frac{680}{r}$ Thus, we have shown that: $A(r) = 2\pi r^2 + 680r^{-1}$

To find the radius that minimizes the surface area, we first compute the derivative of A(r): $A'(r) = 4\pi r - 680r^{-2}$ Setting this equal to zero to find critical points: $4\pi r - \frac{680}{r^2} = 0$ Solving for r, we have: $4\pi r = \frac{680}{r^2}$ Multiplying through by r² results in: $4\pi r^3 = 680$ From this, we solve for r: $r^3 = \frac{680}{4\pi}$ This simplifies to: $r = \sqrt[3]{\frac{680}{4\pi}} \approx 3.78 ext{ cm}$ Thus, we find the radius that minimizes the surface area is approximately 3.78 cm.