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Homework Help: Optimzation problem, only info given are xy intercepts

  1. Nov 19, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose P(a,b) is a fixed point in Quadrant 1 of an xy-plane and line L is descending in the plane such that P is on line L. Let Q=(xknot,0) and R=(0,yknot) be the x and y intercepts for line L and let

    S= 1/(xknot + yknot)

    1. Express S as a function of xknot
    2. Find any extreme values for S

    I know how to do number 2 but number 1 stumps me, how do I get yknot to become xknot?

    2. Relevant equations

    for 2. quotient rule

    3. The attempt at a solution
    so far I have got L=-(y/x) + y using y=mx+b (descending line means it's linear right?)

    in another attempt I find the equation of the line using point RP and then PQ then setting them equal to each other to solve for yknot. I use that to plug into S. S would be expressed in terms of xknot and the a and b are constants. If I find the critical numbers of S with this, it comes out very ugly and does not seem like the answer.

    HOW DO I APPROACH THIS PROBLEM (it is supposedly an optimization problem).
  2. jcsd
  3. Nov 19, 2012 #2


    Staff: Mentor

    It's pronounced like "knot" but the word is "nought", which is synonomous with zero.

    What does it mean that (a, b) is a fixed point? That's key to this problem.
  4. Nov 19, 2012 #3
    I think fixed point P(a,b) means that, plug in "a" into equation L results with "a."
    I get

    S = (x0-a)/x0^2
    Last edited: Nov 20, 2012
  5. Nov 19, 2012 #4
    Am I going in the right direction?
    Last edited: Nov 20, 2012
  6. Nov 20, 2012 #5
    Quick question:

    since x0 and a are points on the graph, is it say to say that the first derivative of S is 0 since they are constants?
  7. Nov 20, 2012 #6


    User Avatar
    Science Advisor

    If [itex]P= (x_0, 0)[/itex] and [itex]Q= (0, y_0)[/itex] then the equation of the line is [itex]x/x_0+ y/y_0= 1[/tex] or [itex]y= y_0- (y_0/x_0)x[/itex]. [itex]x_0[/itex] is NOT a point on the graph it is number- the x-value of the x-intercept. The "derivative of S", since it is a line, is the slope of the line, [itex]y_0/x_0[/itex].

    No, it means that if you put x= a into the equation of the line, you get y= b.
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