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  1. #1
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    Exclamation help with presentation

    Hello I am a thirty something college student who must present a solution to a word problem in class monday. Please help with each step of the solution. Find the length of the shortest ladder that will reach from the ground, over a wall 8ft high, to the side of a building 1 ft behind the wall. That is minimize the legnth L=L1 plus L2
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  2. #2
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    There are various ways to tackle this. I will present a method using trig.

    L=L_{1}+L_{2}

    Then, L=8csc(t)+sec(t) (look at the two triangles made by the ladder, wall, and ground).

    \frac{dL}{dt}=\frac{-8cos^{3}(t)+sin^{3}(t)}{sin^{2}(t)cos^{2}(t)}

    \frac{dL}{dt}=0 if sin^{3}(t)=8cos^{3}(t), \;\ tan^{3}(t)=8

    tan(t)=2 which gives the absolute minimum for L because

    \lim_{t\to\ 0^{+}}L=\lim_{t\to \frac{{\pi}^{-}}{2}}L={\infty}

    If tan(t)=2\Rightarrow csc(t)=\frac{\sqrt{5}}{2} \;\ and \;\ sec(t)=\sqrt{5}

    Hence, L=8(\frac{\sqrt{5}}{2})+\sqrt{5}=5\sqrt{5} \;\ feet
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  3. #3
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    thanks

    is there a way to solve it using derivatives
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  4. #4
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    Quote Originally Posted by galactus View Post
    There are various ways to tackle this. I will present a method using trig.

    L=L_{1}+L_{2}

    Then, L=8csc(t)+sec(t) (look at the two triangles made by the ladder, wall, and ground).

    \frac{dL}{dt}=\frac{-8cos^{3}(t)+sin^{3}(t)}{sin^{2}(t)cos^{2}(t)} *

    \frac{dL}{dt}=0 if sin^{3}(t)=8cos^{3}(t), \;\ tan^{3}(t)=8 *

    tan(t)=2 which gives the absolute minimum for L because

    \lim_{t\to\ 0^{+}}L=\lim_{t\to \frac{{\pi}^{-}}{2}}L={\infty}

    If tan(t)=2\Rightarrow csc(t)=\frac{\sqrt{5}}{2} \;\ and \;\ sec(t)=\sqrt{5}

    Hence, L=8(\frac{\sqrt{5}}{2})+\sqrt{5}=5\sqrt{5} \;\ feet
    Quote Originally Posted by eduk8tedidiot View Post
    is there a way to solve it using derivatives
    What do you think the lines marked with * mean ....!?
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