Maximization

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April 12th 2008, 05:26 PM
topsquark

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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.)

Quote:

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
April 13th 2008, 12:32 AM
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 $\sin\theta = \frac2{15^{1/3}}$ , at which point $y = (15^{2/3}-4)^{3/2}\approx 3.00458$ .
April 13th 2008, 12:34 AM
Moo

Hello,

Isn't it the same as here : http://www.mathhelpforum.com/math-he...n-problem.html ?
April 13th 2008, 04:44 AM
mr fantastic

Quote:

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!
April 13th 2008, 04:54 AM
topsquark

Quote:

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
April 13th 2008, 04:55 AM
topsquark

Quote:

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

Dang it all! I should have seen that. :( Nice work. :)

-Dan