You are designing a 1000 cubic cm closed cylindrical can with radius r and height h. Assume that there is no waste in cutting the aluminum for the side of the can, but that the circular top and bottom are cut from square sheets of aluminum 2r by 2r. Find the radius of the can that minimizes the total amount of aluminum used in manufacturing the can (include the waste from cutting out the top and bottom).
Since we are including the waste for cutting the circles from squares, we need 2 squares and a rectangle
if the radius is r, and our square is 2r by 2r, we can cut 2 circles from 2squares.
Let the height be h , (all units in cm)
we must have:
1000 =π r^2 h
h = 1000/(π r^2)
SA = 2(2r)^2 + 2πrh
= 8r^2 + 2πr(1000/(πr^2)
= 8r^2 = 2000/r
d(SA)/dr = 16r - 2000/r^2 = 0 for a min of SA
16r = 2000/r^2
r^3 = 125
r = 5
So the radius of the can is 5 cm
To find the radius of the can that minimizes the total amount of aluminum used, we need to consider the surface area of the can and the waste from cutting the top and bottom.
Let's start by defining our variables:
- Radius of the can: r
- Height of the can: h
- Surface area of the side of the can: A_side
- Waste from cutting the top and bottom: A_waste
- Total amount of aluminum used: Total aluminum used
The surface area of the side of the can, A_side, can be calculated using the formula for the lateral surface area of a cylinder: A_side = 2πrh.
The waste from cutting the top and bottom, A_waste, is equal to the area of the square sheet minus the area of the circle: A_waste = 2r * 2r - πr^2 = 4r^2 - πr^2.
The total aluminum used, Total aluminum used, can be calculated by adding the area of the side of the can to the waste from cutting the top and bottom: Total aluminum used = A_side + A_waste.
Now, we can substitute the equations for A_side and A_waste into the equation for Total aluminum used:
Total aluminum used = 2πrh + 4r^2 - πr^2.
To find the radius that minimizes the total amount of aluminum used, we need to find the value of r that minimizes this equation. We can do this by taking the derivative of Total aluminum used with respect to r, setting it equal to zero, and solving for r:
d(Total aluminum used)/dr = 0.
Taking the derivative of the equation, we get:
d(Total aluminum used)/dr = 2πh + 8r - 2πr.
Setting this equal to zero and solving for r:
2πh + 8r - 2πr = 0,
6r = 2πh,
r = πh/3.
Therefore, the radius of the can that minimizes the total amount of aluminum used is given by r = πh/3.