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