Find the Length of Shortest Ladder to Reach Over 8 ft. Fence

[MarkFL]

Here is the question:

A fence 8 ft. high (h) runs parallel to a building and is 15 feet (d) from it. Find the length (L) of the shortest ladder that will reach..?

A fence 8 ft. high (h) runs parallel to a building and is 15 feet (d) from it. Find the length (L) of the shortest ladder that will reach from the ground across the top of the fence and to the wall of the building.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Hi,

If we ignore all dimensions of the ladder except the length, this is equivalent to minimizing the sum of the squares of the intercepts of a line passing through the point $(d,h)$ in the first quadrant of the $xy$-plane. Let $a$ and $b$ be the $x$-intercept and $y$-intercept respectively if this line. Thus, the function we wish to minimize is (the objective function):

$f(a,b)=a^2+b^2$

Now, using the two-intercept form for a line, we find we must have (the constraint):

$\displaystyle \frac{d}{a}+\frac{h}{b}=1$

Using Lagrange multipliers, we find:

$\displaystyle 2a=\lambda\left(-\frac{d}{a^2} \right)$

$\displaystyle 2b=\lambda\left(-\frac{h}{b^2} \right)$

and this implies:

$\displaystyle b=a\left(\frac{h}{d} \right)^{\frac{1}{3}}$

Substituting for $b$ into the constraint, there results:

$\displaystyle \frac{d}{a}+\frac{h}{a\left(\frac{h}{d} \right)^{\frac{1}{3}}}=1$

$\displaystyle a=d^{\frac{1}{3}}\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)$

Hence, we have:

$\displaystyle b=h^{\frac{1}{3}}\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)$

and so we find:

$\displaystyle f_{\min}=f\left(d^{\frac{1}{3}}\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right),h^{\frac{1}{3}}\left(d^{\frac{2}{3}}+h^{ \frac{2}{3}} \right) \right)=\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)^3$

Now, we need to take the square root of this since the objective function is the square of the distance we actually wish to minimize. Let $L$ be the length of the ladder, and we now have:

$\displaystyle L_{\min}=\left(d^{\frac{2}{3}}+h^{\frac{2}{3}} \right)^{\frac{3}{2}}$

Letting $d=15\text{ ft}$ and $h=8\text{ ft}$ we have:

$\displaystyle L_{\min}=\left(15^{\frac{2}{3}}+8^{\frac{2}{3}} \right)^{\frac{3}{2}}\text{ ft}\approx32.0134951101408\text{ ft}$