An open-top cylindrical tank with a volume of 2000 cubic meters is to be constructed using steel. The steel used for the top and bottom costs 0.5 dollars per square meter, while the steel used for the sides costs 0.3 dollars per square meter. Determine the dimensions of the tank that will minimize the cost of steel used.
- - take the derivative of this function
- - find the critical points in order to find the maximum and minimum values for your function.
- - prove that the critical points represent maximum or minimum points (eg: with an interval table)
- - find the extreme points on your function
Using this info, and don't use the answer that's already been posted, explain each step and label the answer like extreme points, and use a interval table for number 3
Problem:
An open-top cylindrical tank with a volume of 2000 cubic meters is to be constructed using steel. The steel used for the top and bottom costs 0.5 dollars per square meter, while the steel used for the sides costs 0.3 dollars per square meter. Determine the dimensions of the tank that will minimize the cost of steel used.
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Step 3: State the variables of the problem
Variables:
- r: radius of the base of the cylinder in meters
- h: height of the cylinder in meters
Function: The surface area of the cylindrical tank is given by: A = 2πrh + 2πr2
The cost of steel is given by: C = 0.5(2πr 2) + 0.3(2πrh)
Objective: Minimize the cost of steel C.
Domain:
r and h must be positive numbers. Additionally, since the tank has a volume of 2000 cubic meters, we can use the volume formula for a cylinder to set up an equation relating r and h:
V = πr2h = 2000
Therefore, the domain for r is (0, ∞) and the domain for h is (0, 2000/(πr2)).
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Step 4: Solve the above stated problem
Solution:
Using the equation V = πr2h = 2000, we can solve for h in terms of r: h = 2000/(πr2 )
Substituting h into the cost equation, we get:
C = 0.5(2πr 2) + 0.3(2πr(2000/(πr 2)))
C = πr 2 + 1200/r
Taking the derivative of C with respect to r, we get:
C' = 2πr - 1200/r 2
Setting C' to zero, we get: 2πr = 1200/r 2
Solving for r, we get: r = (600/π)(1/3)
Substituting r back into the equation for h, we get:
h = 2000/(π((600/π)(2/3)))
Therefore, the dimensions of the cylindrical tank that minimize the cost of steel used are approximately:
r = 5.7588 meters h = 19.1961 meters
Thus, the minimum cost of steel used is:
C = π(5.7588) 2 + 1200/19.1961
C ≈ $166.6998
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Solution
The minimum cost of the steel required is
$ 166. 70
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