1. A builder wishes to fence in 60,000 m2 of land in a rectangular shape. For se

Question

1. A builder wishes to fence in 60,000 m2 of land in a rectangular shape. For security reasons, the fence along the front part of the land will cost $20 per meter, while the fence for the other three sides will cost $10 per meter. How much of each type of each fence should the builder buy to minimize the cost of the fence? What is the minimum cost? please show all work.

A) Draw a diagram, label (introduce variables). Write the constraints between your variables, Objective function: write down an expression for the quantity you want to optimize; represent it as a function of one variable, What is the domain of your problem?, Find the minimum cost. Make sure it is clear that you have found the lowest cost, and The builder should buy ( ) meters of $20/m fence and ( ) meters of $10/m fence.  

Explanation / Answer

Solution:

You know you have a fixed area

A = lw

60000 = lw

While the perimeter would normally be

P = 2(l + w)

In this case, we don't want perimeter, instead we want a function of the costs. One length will cost more than the other length and two widths, so we get

C(l, w) = 20l + 10l + 10w * 2

simplifying that

C(l, w) = 30l + 20w

Now we have cost in terms of length and width.

Using the area equation from the above,

solve it for length in terms of width, then substitute it into this equation

60000 = lw

l = 60000/w

C(l, w) = 30l + 20w

C(w) =30(60000/w) + 20w

C(w) = (18,00,000/w) + 20w

Now you have cost in terms of only the width.

To find the minimum cost, solve for the zeroes of the first derivative to get the width of that minimum cost

C'(w) = - (18,00,000/w2) + 20

0 = -(18,00,000 / w2) + 20

Multiply both sides by w2

0 = - 18,00,000 + 20w2

20w2 = 18,00,000

w2 = 90,000

w = 300 m

Now that we have that, we can solve for the length

l = 60000/w

l = 60000/300

l = 200 m

So using that information, we are asked how much fencing is needed of the $20 material? 200 m (one length)

How much of the $10 fencing is needed? (one length and two widths)

= 200 + 300 * 2

= 200 + 600

= 800 m

And in case you were interested, that minimum cost is

C(w) = (18,00,000/w) + 20w

C(300) = (18,00,000/300) + 20*300

C(300) = 6000 + 6000

C(300) = $12,000

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