1. ## Maximization

Here's a problem that's been bugging me. A student asked me this a couple of days ago and I just can't get out of the circles it's twisting.

(See diagram below.)
A 15 ft ladder is constrained to keep one point on the ground. The ladder is propped up by a vertical 8 ft support that can be moved back and forth. What is the maximum horizontal extension of the ladder? (Marked as the line y in the diagram.)
I can't seem to find a way to reduce the possible equations to something with one dependent variable. (This is supposed to be a Calc I problem.) Any takers?

-Dan

2. The length of ladder between the ground and the support point is 8/sin(θ). The length of ladder above the support point is therefore 15 – 8/sin(θ), and the horizontal projection is y = (15 – 8/sin(θ))cos(θ).

The maximum occurs when $\displaystyle \sin\theta = \frac2{15^{1/3}}$, at which point $\displaystyle y = (15^{2/3}-4)^{3/2}\approx 3.00458$.

3. Hello,

Isn't it the same as here : http://www.mathhelpforum.com/math-he...n-problem.html ?

4. Originally Posted by Moo
Hello,

Isn't it the same as here : http://www.mathhelpforum.com/math-he...n-problem.html ?
Nice spot, Moo. And the answer is consistent with the one found using Oplag's suggestion.

Edit: Which is now obvious following Opalg's edit lol!

5. Originally Posted by Moo
Hello,

Isn't it the same as here : http://www.mathhelpforum.com/math-he...n-problem.html ?
That's likely to be because the other poster might be the student I referred to here.

-Dan

6. Originally Posted by Opalg
The length of ladder between the ground and the support point is 8/sin(θ). The length of ladder above the support point is therefore 15 – 8/sin(θ), and the horizontal projection is y = (15 – 8/sin(θ))cos(θ).

The maximum occurs when $\displaystyle \sin\theta = \frac2{15^{1/3}}$, at which point $\displaystyle y = (15^{2/3}-4)^{3/2}\approx 3.00458$.
Dang it all! I should have seen that. Nice work.

-Dan