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How do you solve this optimization problem?

A rancher wants to fence in an area of 600 square feet in a rectangular field using fencing material costing 1.2 dollars per foot, and then divide it in half down the middle with a partition, parallel to one side, constructed from material costing 0.5 dollars per foot.

Assuming that the partition is parallel to the side which gives the width of the field, find the dimensions of the field of the cheapest design.

What is the length? What is the width?

What is the total cost of the cheapest design?

Public Comments

Let A be the area of the field, L the full length and W the width. The area is then A =L*W. The length of the fence is equal to the perimeter of the field which is P = 2*L + 2*W The partition is length W. The total cost of the fencing is then C = P*1.2 + W*0.5. You want to minimize C:

C = 1.2*(2*L + 2*W) + 0.5*W;

you have to get rid of either L or W in the above, so use the known area to get L = A/W then

C(W) = 1.2*(2*A/W + 2*W) + 0.5*W

C(W) = 1/W * (2.4*A + 2.4*W^2 + 0.5*W^2)

C(W) = 1/W * (2.4*A + 2.9*W^2)

Take the derivative of C(W) with respect to W and set it to zero; solve for W; from L=A/W get L and from the cost formula get C
I just took the 400ftx 2 to get the parallel of the rectangle which means the sides would be the 200ft.

Run another fence across it in half would be another 400ft.

The fact that the material was 1.2 dollars seems irrelevant if you want the cheapest design but a rectangular 600ft "field". I guess I don't understand that part of the question.

But anyway, 1200ft + 400ft = 1600ft x .5 = $800

But then again, I failed math.

Thanks for the 2 points!
perimeter=2l+2w

lw=600

w=600/l

P=1200/l+2l

=2(600/l+l)

dP/dl=2(-600/l^2+1)

=2(-600+l^2)/l^2

setting this to zero

l^2=600

l=10rt6

w=10rt6

P=2(l+w)

=2(10rt6+10rt6)

=40rt6

cost=40rt6*1.2

cost of the partition=10rt6*0.5

total cost=$53rt6

=$129.82
area = lxb = 600 or l=600/b

perimeter = 2l + 2b

cost = 2.4l + 2.4b +0.5b

c= 1440/b + 2.9b

dc/db = -1440/b^2 +2.9

for minima dc/db=0

so b = 22.283

to check minima d2c/db2 = 2880/b^3 greater than 0 so minima

l/600/22.283 = 26.926

cost is = 1440/22.283 + 2.9 x 22.283 = 129.24
Let

W = width,

L = length

WL = 600

L = 600/W

C = 1.2*2(W + L) + 0.5W

C = 1.2*2(W + 600/W) + 0.5W

C = 2.9W + 1440/W

dC/dW = 2.9 - 1440/W² = 0

W² = 1440/2.9

W = 22.28344 ft.

L = 600/14.3839 = 26.92582 ft.

C = 2.9*22.28344 + 1440/22.28344

C = $129.243955 = $129.24

check:

2.9*22.2 + 1440/22.2 = 129.24486

2.9*22.3 + 1440/22.3 = 129.24399
The first thing that I did was "prime factorize" 600 ft²:

600 = 2³ * 5² * 3

Now we need to find the possible lengths and widths:

1 * 600 = 600

2 * 300 = 600

3 * 200 = 600

4 * 150 = 600

5 * 120 = 600

6 * 100 = 600

8 * 75 = 600

10 * 60 = 600

12 * 50 = 600

15 * 40 = 600

20 * 30 = 600

24 * 25 = 600

I assume, to get the cost as cheap as possible, that the partition down the middle, would be as short as possible. This means that it will be as long as the shortest side.

cost = [2(l + w) * 1.2] + 0.5w --- This is the perimeter * $1.20 plus the width (shortest side) * $0.50... I am going to simplify the equation by combining like terms.

cost = 2.4l + 2.9w

Now all we need to do is plug in or derived lengths and widths:

2.4(25) + 2.9(24) = $129.60

2.4(30) + 2.9(20) = $130.00

2.4(40) + 2.9(15) = $139.50

2.4(50) + 2.9(12) = $154.80

2.4(60) + 2.9(10) =

2.4(75) + 2.9(8) =

2.4(100) + 2.9(6) =

2.4(120) + 2.9(5) =

2.4(150) + 2.9(4) =

2.4(200) + 2.9(3) =

2.4(300) + 2.9(2) =

2.4(600) + 2.9(1) =

Answer: Length = 25 feet, Width = 24 feet, and the cost is $129.60.

~ Note: I got to thinking when I was thinking of the pairs, "What about the square root of 600?"

√600 ≈24.49

This would make the length 24.49 feet and the width 24.49 feet.

2.4(24.49) + 2.9(24.49) = $129.80 --- $0.20 more.

Just thought I'd share. ~