1. The problem statement, all variables and given/known data A cylindrical can is to hold 4π cubic units of juice. The cost per square unit of constructing the metal top and the bottom is twice the cost of constructing the cardboard side. what are the dimensions of least expensive can? calculus must be used logically to solve this problem and work should be shown. 2. Relevant equations The total surface area of the can is: A = 2πrh + 2πr² The volume of the can is: V = πr²h = 4π, then h = 4/r² 3. The attempt at a solution connect volume and area equation, A and V. so that; A = 2πr(4/r²) + 2πr² We want the area of the top to be a minimum while having the same volume. Therefore, we need to minimize r and vary the height. We will take the derivative of the total surface to accomplish this aim. dA/dr = 8π(-1/r²) + 4πr = 0, r = ³√2, and since the cost of Area of top is twice the area of cardboard; that is, 2πr² = 2(2πrh), then h = r/2 The dimensions for minimum material cost is then: r = ³√2, and h = 2 ³√2 I did untill this. And I will plug this h value and r value in area equation. But I am nut sure i am doing right. especially the part with [since the cost of Area of top is twice the area of cardboard;] Pleas give me a hand.