1. ## 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

2. ## Re: Calculus - Maximum length of ladder

I'm not sure what happened on the last step: You have $\displaystyle \frac{64\sin{x}}{\cos^2{x}}-\frac{48\cos{x}}{\sin^2{x}}=0$, which should give you:

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

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

- Hollywood

3. ## 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

4. ## 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):

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

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

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

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

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

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:

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

- Hollywood

5. ## 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 $\displaystyle a$ and $\displaystyle b$ and I thought you might find it interesting. It's attached to my reply.