# Calculus - Maximum length of ladder

• Apr 5th 2013, 07:24 AM
MMCS
Calculus - Maximum length of ladder
Attached is my working of the current problem, but i get an invalid solution. Any help would be appreciated
Thanks
• Apr 5th 2013, 08:02 AM
hollywood
Re: Calculus - Maximum length of ladder
I'm not sure what happened on the last step: You have $\frac{64\sin{x}}{\cos^2{x}}-\frac{48\cos{x}}{\sin^2{x}}=0$, which should give you:

$\frac{48}{64}=\tan^3{x}$, so x is approximately 0.7375 radians or 42.3 degrees.

That gives you a ladder length of $\frac{48}{\sin{x}}+\frac{64}{\cos{x}}=157.85$.

- Hollywood
• Apr 5th 2013, 08:39 AM
MMCS
Re: Calculus - Maximum length of ladder
thanks for your reply hollywood, Is it possible for you to exaplain how you got from what i had to what you have as i am new to calculus
Thanks
• Apr 5th 2013, 06:04 PM
hollywood
Re: Calculus - Maximum length of ladder
Sure, but you already did the hard part...

Starting with your equation (recall you took the derivative and set it equal to zero):

$\frac{64\sin{x}}{\cos^2{x}}-\frac{48\cos{x}}{\sin^2{x}}=0$

add $\frac{48\cos{x}}{\sin^2{x}}$ to both sides:

$\frac{64\sin{x}}{\cos^2{x}}=\frac{48\cos{x}}{\sin^ 2{x}}$

multiplly both sides by $\frac{\sin^2{x}}{64\cos{x}}$ and combine the trig functions:

$\tan^3{x}=\frac{48}{64}$

Then take the cube root and inverse tangent to get x=0.7375. But you want the length of the ladder, so you have to plug in to the original equation:

$\text{length}=\frac{48}{\sin{0.7375}} + \frac{64}{\cos{0.7375}}=71.38+86.47=157.85$.

- Hollywood
• Apr 7th 2013, 12:37 AM
nimon
Re: Calculus - Maximum length of ladder
Interesting, I was just doing this problem in a book I'm reading. The solution I found doesn't use the angle at all, but instead labels the lengths with new variable names and finds a relationship between them in terms of the known variables. The answer is given in terms of $a$ and $b$ and I thought you might find it interesting. It's attached to my reply.