Longest Beam Problem

turophile

Junior Member
Joined
May 22, 2010
Messages
94
The problem:

A beam must be carried around a corner from a hall of width A to a hall of width B. What is the length of the longest beam that will turn the corner?

What I've done so far:

Let the length of the portion of the beam that extends into the hall of width A be P, let the length of the portion that extends into the hall of width B be Q, and let the angle that the beam forms with the inside wall of the hall of width B as it turns the corner be ?. Let L be the beam length (L = P + Q).

sin ? = B/Q ? Q = B/(sin ?)

sin (?/2 – ?) = cos ? = A/P ? P = A/(cos ?)

L = A sec ? + B csc ?

d/d? (A sec ? + B csc ?) = A sec ? tan ? – B csc ? cot ? ? 0

? A ? (sin ?)/(cos[sup:3pnwojev]2[/sup:3pnwojev]?) = B ? (cos ?)/(sin[sup:3pnwojev]2[/sup:3pnwojev]?)

? (sin[sup:3pnwojev]3[/sup:3pnwojev]?)/(cos[sup:3pnwojev]3[/sup:3pnwojev]?) = tan[sup:3pnwojev]3[/sup:3pnwojev]? = B/A

? tan ? = B[sup:3pnwojev]1/3[/sup:3pnwojev]/A[sup:3pnwojev]1/3[/sup:3pnwojev] = (B/Q)/(A/P) = (BP)/(AQ)

? B[sup:3pnwojev]1/3[/sup:3pnwojev]AQ = A[sup:3pnwojev]1/3[/sup:3pnwojev]BP

? A[sup:3pnwojev]2/3[/sup:3pnwojev]Q = B[sup:3pnwojev]2/3[/sup:3pnwojev]P

? Q = (B/A)[sup:3pnwojev]2/3[/sup:3pnwojev]P

L = P + Q = P + (B/A)[sup:3pnwojev]2/3[/sup:3pnwojev]P = P[1 + (B/A)[sup:3pnwojev]2/3[/sup:3pnwojev]]

My question:

I'm stuck here and can't see how to solve for L in terms of A and B. Any hints?
 
It looks like you're close to the general solution.

Let a and b be the width of the halls and the length of the ladder be L=p+q.

We could use similar triangles or trig. Since you are using trig, I will go with that.

I will start here:

\(\displaystyle tan\theta=\frac{b^{\frac{1}{3}}}{a^{\frac{1}{3}}}\)

\(\displaystyle sec\theta=\frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}\)

\(\displaystyle csc\theta=\frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}\)

\(\displaystyle L=\underbrace{a\cdot\frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{a^{\frac{1}{3}}}}_{\text{p}}+\underbrace{b\cdot \frac{\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}}_{\text{q}}\)

\(\displaystyle L=a^{\frac{2}{3}}\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}+b^{\frac{2}{3}}\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\)

Factor:

\(\displaystyle L=\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}(a^{\frac{2}{3}}+b^{\frac{2}{3}})\)

\(\displaystyle \boxed{L=(a^{\frac{2}{3}}+b^{\frac{2}{3}})^{\frac{3}{2}}}\)
 
As I mentioned, this can be done with similar triangles.

\(\displaystyle \frac{q}{p}=\frac{p}{\sqrt{p^{2}-a^{2}}}\)

\(\displaystyle q=\frac{bp}{\sqrt{p^{2}-a^{2}}}\)

\(\displaystyle L=p+\frac{bp}{\sqrt{p^{2}-a^{2}}}\)

Differentiating, setting to 0, and solving for p gives:

\(\displaystyle p=a^{\frac{2}{3}}\sqrt{a^{\frac{2}{3}}+b^{\frac{2}{3}}}\)

q can be found by subbing in above, then adding p+q=L.

Same result.
 
Top