-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 the info below 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.arrow_forwardStep 3: State the variables of the problemVariables:r: radius of the base of the cylinder in metersh: height of the cylinder in metersFunction: The surface area of the cylindrical tank is given by: A = 2πrh + 2πr2The 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 = 2000Therefore, the domain for r is (0, ∞) and the domain for h is (0, 2000/(πr2)).arrow_forwardStep 4: Solve the above stated problemSolution: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/rTaking the derivative of C with respect to r, we get:C' = 2πr - 1200/r 2Setting C' to zero, we get: 2πr = 1200/r 2Solving 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 metersThus, the minimum cost of steel used is:C = π(5.7588) 2 + 1200/19.1961C ≈ $166.6998 arrow_forwardSolutionThe minimum cost of the steel required is$ 166. 70and the rest are attached on the photo Step 1: GivenVolume of cylindercal tank = 2000 m³ Let radius = rm, height = h mStep 2: Explanationr²h = 2000 ⇒ h =2000π²...iVolume of cylindrical tank = 2000Cost of top and bottom 1 m² area = $0.5Total cost of top and bottom = 0.52nr² = r² Area of sides = 2πrhСos oTotal cost of cylinder = C ⇒ C = r² +0.6h ⇒ Сx = r² +0.6² ⇒ Cx = : Tr² +12002000π²rTake derivative with repect to r1200⇒ C'x = =nr² +dr2πr -⇒ С'x =Put C'(x)=0⇒ C"x = 2n-1200-2²⇒ C"x = 2π-1200-2r-²-12400⇒ C"x = 2πt +3Atr=120012006003⇒ C²x = 2πr - = 0⇒ 2² = 12 ⇒ 2m³ = 1200⇒ r =Now take second derivative with respect to r of C(r)d1200⇒ C"x=== 2πr-drdr6003Ar r=⇒ C"x = 2π +Solution120016003drπ2400600TTdrdx5.7588 m Height h> 0 So cost is minimum at r=ConclusionHenceMinimize cost of steel is $312.56= nx²-120006001600 3TI'm19.1961 mMinimum cost = ² +1200r6003⇒ С= n°TI+1200600ī$312.56nearest

1. volume of open-top cylindrical tank V=2000 cu meter 2. Cost of steel used for top and bottom is…

Answered: An open-top cylindrical tank with a…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Related questions

Question

-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 the info below

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.

arrow_forward

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)).

arrow_forward

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