-Analyze the extreme points and critical points to determine the optimum value that solves your problem. -end your problem with a concluding statement  -find the equation of the tangent at your optimum point  -create a graph that presents: your function, the derivative and the tangent at the optimum point. Using this info, and don't use the answer that's already been posted, explain each step and label the answer   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.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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  •  -Analyze the extreme points and critical points to determine the optimum value that solves your problem.
  • -end your problem with a concluding statement 
  • -find the equation of the tangent at your optimum point 
  • -create a graph that presents: your function, the derivative and the tangent at the optimum point.

Using this info, and don't use the answer that's already been posted, explain each step and label the answer  

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

 

The minimum cost of the steel required is $312

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