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Homework Help: Maximizing area word problem WITHOUT LP formula

  1. Jan 29, 2012 #1
    The problem statement, all variables and given/known data
    We have to solve this problem using only what we learned in calculus (no linear programming)
    I attached a picture to help :)

    A farmer wants to build a rectangular pen. He has a barn wall 40 feet long, some or all of which must be used for all or part of one side of the pen. In other words, with f feet of of fencing material, he can build a pen with a perimeter of up to f+40 feet, and remember he isn't required to use all 40 feet.
    What is the maximum possible area for the pen if 100 feet of fencing is available?


    The attempt at a solution
    I attempted to solve by finding the critical points of the derivative:
    2x+y=100 --> y=100-2x
    x(100-2x)=100x-2x^2
    f'(x)=100-4x
    100-4x=0 --> x=25
    2(25)+y=100 --> y=50
    A=1250

    BUT I realize this is wrong because if y=50, then the side of the barn needs an extra 10 feet, but all of the fencing material has been used by the other 3 sides. Help please?
     

    Attached Files:

    Last edited: Jan 29, 2012
  2. jcsd
  3. Jan 29, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You can write this problem as
    maximize x*y
    subject to
    x + 2*y <= 100,
    x <= 40,
    x,y >= 0.

    It is a nonlinear programming problem, because of the inequality constraints. If you haven't studied such problems yet, you should note that you can reduce it to 2 cases: (1) x + 2y < 100; (2) x + 2y = 100. In case (1) the constraint 2x+y <= 100 is not active, so it is like an unconstrained problem (but with a bound on x). In this case the y-derivative of A = x*y should be zero at an optimum, and doing that gives x = 0. This is obviously not a maximum, since we can do better than area = 0. So, we need to be in case (2): use all the available fencing. Thus, y = 50 - x/2, and putting that into A = x*y we have a function f(x) = x*(50 - x/2) to be maximized over 0 <= x <= 40.

    As you have realized, you can't just set f'(x) = 0 because that violates the bound of 40. But *think about it*: why would you want to set the derivative of f to zero? You would want to to that because for x less than the critical point f is increasing, and after the critical point f is decreasing. In this case the critical point lies outside the allowed region, so what does that tell you about the max allowed value? Think about plotting a graph of the function f(x).

    RGV
     
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