1. ## Optimization

One hallway which is 4 feet wide meets another hallway which is 8 feet wide in a right angle (see picture). What is the length of the longest ladder which can be carried horizontally around the corner? (Hint: We need to minimize the length of the ladder so it just fits as shown in the picture. Express the length in terms of the angle θ shown.)

I couldn't copy the picture properly. Sorry for that.

2. Did you know there is a general formula for these 'hallway' problems?.

$\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\fra c{3}{2}}$

But, we can derive yours and show the formula works.

Using theta is a hint, not a requirement?.

When this problem gets interesting is when the width or diameter of the object is taken into account. This one is negligible though.

The length of the ladder is given by L=x+y

By similar triangles:

$\frac{y}{8}=\frac{x}{\sqrt{x^{2}-16}}, \;\ y=\frac{8x}{\sqrt{x^{2}-16}}$

$L=x+\frac{8x}{\sqrt{x^{2}-16}}, \;\ for \;\ x>4$

$\frac{dL}{dx}=1-\frac{128}{(x^{2}-16)^{\frac{3}{2}}}$

set to 0 and solve for x.

You will find the answer will be the same if you plug in a=4 and b=8 into the formula I gave at the beginning.

3. Thanks! I think I understand your process. I do know that expressing the equation with θ is a requirement though and I'm not sure how I could do that. I would have to use Trig, right?

I finally got my sister to make a picture for me, if it'll help:

4. Originally Posted by ebonyscythe
Thanks! I think I understand your process. I do know that expressing the equation with θ is a requirement though and I'm not sure how I could do that. I would have to use Trig, right?

I finally got my sister to make a picture for me, if it'll help:
At the 4-ft wide hallway.
Let x = portion of the ladder there
LetT = theta

sinT = 4/x
x = 4/sinT ------------**

At the 8-ft wide hallway,
sin(90-T) = 8/(L-x)
cosT = 8/(L-x)
L-x = 8/cosT ----------**

So,
L -4/sinT = 8/cosT
L = 4/sinT +8/cosT -------------------------(i)
L = 4(sinT)^(-1) +8(cosT)^(-1) --------------(ii)
Differentiate (ii) with respect to angle T,
dL/dT = -4cosT/(sin^2(T)) +8sinT/(cos^2(T))
Set that to zero,
0 = -4cos^3(T) +8sin^3(T)
8sin^3(T) = 4cos^3(T)
sinT = cubrt(1/2) *cosT
tanT = cubrt(1/2)
T = arctan[cubrt(1/2)] = 38.439 degrees

Therefore,
minimum L = 4/sin(38.439deg) +8/cos(38.439deg) = 16.647 ft. ---answer.

That is for the "maximum" length of the ladder.

------------------
minimum L?

If T = 38deg, L = 16.649'
T = 38.439deg, L = 16.647'
If T = 39deg, L = 16.650'
So graph goes to the bottom and then goes up again, hence 16.647' is minimum.

5. To do it in terms of angles, you must know how to use your trig functions.

$L=4sec{\theta}+8csc{\theta}$

Now, differentiate:

$\frac{dL}{d{\theta}}=4sec{\theta}tan{\theta}-8cot{\theta}csc{\theta}$

Set to 0 and solve.

It simplifies down to:

$4sec{\theta}tan{\theta}=8cot{\theta}csc{\theta}$

$\frac{1}{2}=cot^{3}{\theta}$

${\theta}=tan^{-1}(2^{\frac{1}{3}})\approx{0.899908348082}$

Plug that back into the Length formula to get your length. It shopuld be the same as the other method.

$4sec(.899908...)+8csc(.899908...)=16.65$

6. How does: $1 - \frac{ba^2}{(x^2 - a^2)^{3/2}}$ become x = $\left(a^{\frac{2}{3}}+b^{\frac{2}{3}}\right)^{\fra c{3}{2}}$ ?

I can only get $1 - \frac{ba^2}{(x^2 - a^2)^{3/2}} =$ $x = \sqrt{a^2 + ba^{4/3}}$

7. If the length we get, 16.65 is a minium then how is that the largest possible ladder. Is it because we have to minimize theta in order to maximize length.. but does that make sence... the larger theta is the larger the length would become rite?