Find the maximum horizontal overhang for a 15 foot ladder over an 8 foot fence
The ladder base is always on the ground, the fence is vertical and ground it horizontal
1. Draw a sketch. Let x denote the base of the ladder and y the overhang.
2. Use proportions:
$\displaystyle \frac8x = \frac z{x+y}~\implies~z=\frac{8(x+y)}{x}$
3. Use Pythagorean theorem:
$\displaystyle (x+y)^2 + z^2=15^2~\implies~(x+y)^2+\left(\frac{8(x+y)}{x}\r ight)^2=225$
4. Solve for y. That means: Calculate the overhang with respect to the base. You'll get a positive and a negative result. The negative result isn't very plausible here. Thus:
$\displaystyle y = \frac{15x}{\sqrt{x^2+64}}-x$
5. Calculate the first derivative (it's easier (for me!) to use product rule and chain rule):
$\displaystyle y' = \frac{960}{\left(\sqrt{x^2+64}\right)^3}-1$
6. Solve for x: y' = 0. I've got $\displaystyle x = 4 \cdot \sqrt{\sqrt[3]{225}-4} \approx 5.77193...$
Let F denote the foot of the ladder which is moved horizontally.
Then the top of the ladder describes a curve which I have "recorded". The vertical tangent to this curve indicates the maximum overhang.
According to my previous calculations the overhang is y = 3.004583...
This answer is consistent with that found using the method suggested by Opalg at http://www.mathhelpforum.com/math-he...imization.html.
Well, on the other hand:
1. You always have lovely graphics (it really is your trademark)
2. You reply is dated 11 March 11:20 am and Oplag's is dated 14 March 12:32 am. So it took Opalg 3 more days than you to get the elegant solution Lives might have been lost if not for your less elegant but timely solution
(ps Only kidding, opalg )