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Homework Help: Optimization Help!

  1. Nov 17, 2011 #1
    So I have a problem. The problem says:
    A fence 8 ft tall rubs parallel to a tall building at a distance of 4 ft from the building. What is the length of the shortest ladder that will reach from the ground over the fence to the wall of the building?

    I drew a picture to help, but I can't draw it on here. I drew a right triangle, it's fairly large. Halfway through the base, there's a straight line up representing the fence. I then made the hypotenuse of the entire triangle L. The wall is h.

    1.I then set up a proportion. It's (x/8)=((x+4)/(h))
    2.I solved the proportion for x and got 32/(h-8)
    3.I then said (x+4)^2+h^2=L^2
    I plugged in the x value i solved the proportion for into the above equation and got
    (((4h)/(h-8))^2)+(h^2)^(1/2))=L (It might help to right this out instead of looking at it.

    Now I have to take the derivative of this and set it equal to zero.. But I keep messing it up.. Can somebody show me how to do it?

    I have been working and realized I can take the derivative of the Pythagorean theorem mentioned in step 3. But I still have trouble solving this. can somebody please help?
    Last edited: Nov 17, 2011
  2. jcsd
  3. Nov 19, 2011 #2


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    This problem is a lot easier if you use trig functions. Start with 4+x = L*cos(θ) and write an expression for x in terms of θ using the small triangle. Substitute and differentiate L with respect to θ. It works out quite nicely.
  4. Nov 19, 2011 #3


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    I just did this problem with a student this past week (Stewart, Section 4.7, no?). The trick if you do this algebraically is not to simplify your expression too soon. It will also be better if you eliminate h , rather than x . Your Pythagorean result becomes

    [tex]L^{2} = (x + 4)^{2} + (\frac{8 \cdot (x+4)}{x})^{2} . [/tex]

    There are two things that will be helpful to do. One is to factor out the ( x + 4 )2. The other is to differentiate the expression for L2 implicitly with respect to x . When you are looking to minimize a length given by the Pythagorean Theorem or the distance formula, since that length is a positive number, minimizing L2 also minimizes L . You avoid dealing with a radical until you absolutely have to (if you need to at all).

    We will then have

    [tex] \frac{d}{dx} L^{2} = \frac{d}{dx} [ (x + 4)^{2} [ 1 + \frac{64}{x^{2}}] ] = 0 [/tex]

    [tex]\Rightarrow 2L \cdot \frac{dL}{dx} = 2 \cdot (x + 4) ( 1 + \frac{64}{x^{2}}) + [ (x + 4)^{2} ( - \frac{128}{x^{3}}) ] = 0 [/tex]

    Since L is not equal to zero , we can just focus on getting the right-hand side of this equation to be zero. Also, x ≠ -4 , so we can divide that factor out of the equation; we can also take out a factor of 2 . This will leave

    [tex] ( 1 + \frac{64}{x^{2}}) + [ (x + 4) ( - \frac{64}{x^{3}}) ] = 0 \Rightarrow 1 + \frac{64}{x^{2}} = (x + 4) ( \frac{64}{x^{3}}) \Rightarrow 1 + \frac{64}{x^{2}} = \frac{64}{x^{2}} + \frac{256}{x^{3}} \Rightarrow \frac{256}{x^{3}} = 1 . [/tex]

    You can take it from there (I'll tell you that the minimized length is ≈16.65 ft.)

    The trigonometric method hotvette describes is the one the solver for Stewart's solution manual used. It is "easier" in the sense that the differentiation to find the critical value for [itex]\theta[/itex] is simpler; you pay for it later in extracting the minimal length for the ladder (particularly if you are solving for the "exact" value). You get a different but equivalent expression for the length, which is confirmed when you calculate the decimal approximation.
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