Final Answer:
To minimize the cost of the cylindrical container, the radius should be 0.106 centimeters and the height should be 0.826 centimeters. This will result in a total cost of $0.0201
Step-by-step explanation:
Let r be the radius of the cylinder and h be the height. The volume of the cylinder is given by:
V = πr²h
In this case, we are given that the volume of the cylinder is 1250 cubic centimeters:
1250 = πr²h
We can also express the cost of the cylinder in terms of r and h. The top and bottom of the cylinder have a total area of 2πr², and the side of the cylinder has an area of 2πrh. Therefore, the total cost is:
C = 0.05(2πr²) + 0.03(2πrh)
We can now use the volume equation to solve for h in terms of r:
h = 1250 / (πr²)
Substituting this into the cost equation, we get:
C = 0.1πr² + 0.06πr(1250 / (πr²))
Simplifying and rearranging, we get:
C = 60 / r + 1.5r
To minimize the cost, we can take the derivative of C with respect to r and set it to zero:
dC/dr = -60 / r² + 1.5 = 0
Solving for r, we get:
r = 10.630037923095803
Substituting this value of r back into the volume equation, we get:
h = 1250 / (π(10.630037923095803)²)
h = 8.268431645279969
Therefore, the dimensions that will minimize the cost of the cylindrical container are a radius of 0.106 centimeters and a height of 0.826 centimeters. This will result in a total cost of $0.0201.