# Thread: Optimization Problem

1. ## Optimization Problem

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

2. Originally Posted by Exiab
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 $\displaystyle x = 4 \cdot \sqrt{\sqrt[3]{225}-4} \approx 5.77193...$

3. 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...

4. Originally Posted by earboth
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.

5. Originally Posted by mr fantastic
This answer is consistent with that found using the method suggested by Opalg at http://www.mathhelpforum.com/math-he...imization.html.
Opalg's method is much more elegant than mine. Unfortunately such solutions never occur to me ...

6. Originally Posted by earboth
Opalg's method is much more elegant than mine. Unfortunately such solutions never occur to me ...
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 )