.Step 1:
The material forming the top and bottom is two times the area of a circle with a radius r. SA = 2(SA) = 2π r²
The material forming the side is equal to height h times the circumference of a circle with radius r.
SA = 2πrh
Therefore, the material needed in the can is M = 2π r² + 2πrh.
However, h is independent of r: if r grows, h shrinks, and vice versa. To find the relationship between h and r, use the volume of a cylinder formula (V = π r²h) because the volume of the can is a constant 40 in³.
40 = π r²h, giving h = .
This means that the material for the side = 2π rh = 2π r = 
.
We obtain the formula for the Total Material in the can in terms of radius r:
M(r) = 2π r² + .
Step 2:
The domain of the function is r > 0.Step 3:
The graph of M(r) = 2π r² + 80/r is shown with M(r) = 64.7 below.
It also shows the trend line with the radius being the domain and the total material used being the range. The graph confirms the data in the above chart. It shows that when the radius gets very small or very large, the material usage increases. It appears that the low point of material usage is within [1.0, 3.0].
Step 4:
Now use calculus to find the minimum of M. Find the derivative of M(r). 
0 =
4πr = 
πr³ = 20
r = 
≈ 1.85 inches
Therefore, the length of radius that minimizes the amount of material for each can is approximately 1.85 inches. This agrees with our interval of [1, 3] and the graph of the function.
Step 5:
Find h by substitution. h =˜ 
≈ 3.7 inches
Conclusion: The size of can that will hold 40 in³ of juice and will minimize use of aluminum has a radius of approximately 1.85 inches and a height of 3.7 inches. This size cylinder will need approximately 64.7 in² of aluminum.