To solve this problem let us say that, h = height of the can r = radius of the can We try to minimize the amount of metal used which is the surface area (SA), and the equation for a cylinder is: SA = 2πrh + πr^2 To get the minima, we take the derivative of this function and set it equal to 0. But first, the function is in two variables so we must eliminate one of them. We use the extra information given in the problem which is volume: V = 8π = πr^2 h Therefore, h = 8/r^2 Plug this into the SA function to have it in terms of one variable only: SA = 2πrh + πr^2 SA = 2πr(8/r^2) + πr^2 SA = 16π/r + πr^2 Taking the 1st derivative of the function: 0 = −16π r^−2 + 2πr 0 = −16π/r^2 + 2πr 16π/r^2 = 2πr r^3 = 8 r = 2 in By obtaining r, we calculate for h: h = 8/r^2 h = 8/(2)^2 h = 8 / 4 h = 2 in (ANSWER)
A cylindrical metal can is to have no lid. it is to have a volume of 8Ï in3. what height minimizes the amount of metal used?