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Homework Help: Use of differentiation in order to find the minimium value

  1. Jul 9, 2006 #1
    The owner of a garden centre wishes to fence a rectangular area of 300 meters squared. She wished to fence three sides with a fencing that costs $9 per meter and the forth side with fencing costing $15 per meter. Find the dimensions of the rectangular area that will minimise her fencing cost.

    I have started off this problem by drawing out a rectangle which I have labelled 2 sides as x and the other 2 as y. then I found y in terms of x according to the set area which was given. However I am not sure on how to proceed with this problem, as I am confused on how to find a function for the cost of this fencing. I know that this problem requires the use of differentiation in order to find the minimium value.

    Thanks to anyone who helps and all input welcome :smile: ,
  2. jcsd
  3. Jul 9, 2006 #2


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    What are you confused about? The costs are found by multiplying the cost per meter by the length. Add up the costs of each side and minimize with respect to x (the area relation should allow you to eliminate y).
    Last edited: Jul 9, 2006
  4. Jul 10, 2006 #3


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    Three sides of the fence are the less expensive $9 per foot material and one side is $15 per foot. You will need to decide whether it is an "x" or "y" side that has the more expensive material!

    Suppose you choose "y" to represent the length with the $15 per foot material. How much would that cost altogether? Now you have 2 sides of length "x" and 1 side of length "y". How long is that altogether? How much will that cost at $9 per foot?
  5. Jul 11, 2006 #4
    thanks for the replies
    i understand that i need to label the sides as having different variables, but when i do that i get an answer of [tex]3\sqrt5000[/tex] for x, and where [tex]y = 300 - 3x[/tex], however this in incorrect because the answer in the book states that the best design for this enclosure would be 15m x 20m where one of the 15m lengths would be the $15m fencing. I do not understand how they conculded to this answer. many thanks,
  6. Jul 11, 2006 #5
    The problem can be restated as

    Min J = 9x + 9x + 9y + 15y = 6*(3x+4y)

    Subject To x*y = 300

    This problem is a direct optimization problem meaning it does not require the use of Lagrange mulitpliers to develop a solution.

    We can simply substute the constraint equation into the cost function

    that is y = 300/x

    so we now have min J = 6*(3x+1200/x)

    we can also throw out the 6 for determining the value of x however if we want to compute an exact cost we must include it

    so now you problem is

    minimize the function J = 3x + 1200/x obviously we must only use x > 0

    I think you can do the rest, to find the value of y knowing x, just simply apply the constraint equation

    Also when I say cost function that is the line that says Min J = f(x,y)
    the constraint equation is A = x*y

    Its interesting to notice that there exists a plane of closed form solutions to the problem for variations of area and the ratio of the cost of the 3rd side to the remaining sides.

    As a check for your problem make sure you obtain

    x = sqrt(0.5*(1+b)*A)

    where b is the ratio of the 3rd side to the remaining sides
    and A is the area
    Last edited: Jul 11, 2006
  7. Jul 11, 2006 #6


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    In other words, it was the part you originally said you could do that causes the problem! The area is xy= 300 so y= 300/x, NOT y= 300- 3x.
    (The only way I can see that you would get that is if you thought that 3 of the sides were of length x and used the perimeter formula rather than area.) There are 2 sides of length x and 2 sides of length y. If we make x the length of the side that cost $15 per foot, the cost of that side is 15x. The total length of the other 3 sides is 2y+ x and that cost $12 per foot. What is the cost of those 3 sides? What is the total cost function?
  8. Jul 11, 2006 #7
    many thanks for the replies. i now understand how to go about this problem and my approach from before was wrong. thanks once again,
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