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anzhelika [568]
4 years ago
13

A fenced enclosure consists of a rectangle (length L and width 2R) attached to a semicircle with a radius R as pictured below. N

ote: there is fencing between the rectangular and semi-circular portions. The enclosure is to be built to have a total area (the entire shaded region), A, of 2000 ft2. The cost of the fence is $20/ft for curved sections and $30/ft for straight sections. Analytically (show all of your steps using algebra and calculus in the write up), find the minimum cost to build the fence and the dimensions of the enclosure. Show all equations that you derive and use. This part should be done entirely in the write-up: no coding. It is fine to do this by hand and include a high-resolution picture in the writeup rather than using Equation Editor. Hint: you will need to use two equations in order to find the cost as a function of only R or L. Construct a plot to graphically determine the values of R and L (x-axis) that minimize the total cost (y-axis) of the fence. Using the MIN function, determine the R and L that minimize cost and what that cost is Do your answers in A, B and C all agree? Why or why not?

Mathematics
1 answer:
Romashka [77]4 years ago
8 0

Answer:

A. $ 6053.44

B. From the graph, R = 21.027 ft and C = $ 5700.005, L = 31.043 ft

The values in A and B do not all agree. This could be due to error in approximations. Their values are close though.

Step-by-step explanation:

A. The area A of the enclosure equals, A = 2RL + πR²/2.

The total cost C = 20 × length of curved section + 30 × length of straight section

C = 20πR + 30[2(L + 2R)]

   = 20πR + 60L + 120R

Making L subject of the formula from A,

L = A/2R - πR/4

Substituting L into C, we have

C = 20πR + 60(A/2R - πR/4) + 120R

   = 20πR + 30A/R - 15πR + 120R

   = 5πR + 30A/R + 120R

We now differentiate C with respect to R to find the value of R for minimum cost

dC/dR = 5π - 60A/R² +120

Equating dC/dR to zero, we have

5π - 60A/R² + 120 = 0

So, R = ±√[60A/(5π + 120)]

substituting A = 2000 ft²

R = ±√[60 × 2000/(5π + 120)] = ±29.74 ft

We take the positive answer, R = 29.74 ft since R cannot be negative.

To determine if this is a minimum point, we differentiate dC/dR with respect to R.

So d²C/dR² = 120A/R³

Since d²C/dR² = 120A/R³ > 0 for positive R, it is a minimum point.

Substituting the value of R into C we have

C = 5πR + 30A/R + 120R

   = 5π(29.74) + 30 × 2000/29.74 + 120(29.74)

   = 467.155 + 2017.485 + 3568.8

   = $ 6053.44

and L = A/2R - πR/4

         = 2000/2(29.74) - π(29.74)/4

         = 33.625 - 23.356

         = 10.269

         ≅ 10.27 ft

B. From the graph, R = 21.027 ft and C = $ 5700.005, L = 31.043 ft

The values in A and B do not all agree. This could be due to error in approximations. Their values are close though.

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