# Maximization problem

I've been trying to solve the following problem from Stewart's Calculus Textbook for a while without any success. My answer makes sense, but I'm looking for a way to solve it analytically.

The problem concerns a pulley that is attached to the ceiling of a room at a point C by a rope of length r. At another point B on the ceiling, at a distance d from C (where d > r), a rope of length l is attached and passed through the pulley at F and connected to a weight W. The weight is released and comes to rest at its equilibrium position D. This happens when the distance |ED| is maximized. Show that when the system reaches equilibrium, the value of x is:

$$\frac{r}{4d}(r+\sqrt{r^2+8d^2})$$

Here is what I've done. First, I expressed |DE| as a function of x

$$|DE|(x)={a}_{2}+{a}_{3}=l-{a}_{1}+\sqrt{{r}^{2}-{x}^{2}}=l-\sqrt{{a}_{3}^2+y^2}+\sqrt{r^2-x^2}$$

from what follows that $$|DE|(x)=l-\sqrt{r^2+d^2-2xd}+\sqrt{r^2-x^2}$$ defined for $$0\leq x \leq r$$

...and it works since $$|DE|(0)=l+r-\sqrt{r^2+d^2}$$ and $$|DE|(r)=l-|r-d|$$

To find the maximum of this function, I calculated |DE|'(x)

$$|DE|'(x)=\frac{d}{\sqrt{r^2+d^2-2xd}}-\frac{x}{\sqrt{r^2-x^2}}$$

I proved the two radicals at the denominator are defined for $$0\leq x < r$$

...so basically I'm interested in finding when |DE|'(x) equals zero, more specifically the roots of

$$d\sqrt{r^2-x^2}-x\sqrt{r^2+d^2-2xd}=0$$ that becomes

$$2dx^3-(r^2+2d^2)x^2+d^2r^2=0$$

I graphed |DE|(x) and |DE|'(x) (using l = 15, r = 3, and d = 4), they are consistent with the problem. |DE|'(x) has only one root at about 2.76 and at the same point |DE|(x) has its maximum. Moreover, if you substitute my test numbers for l, r, and d in the given formula for |DE|(x) maximum you get the same numerical result.

So, how was the author of the problem able to find an analytical solution to the problem?

Thanks!

• I appreciate the diagrams, $TeX$, and work. This is a great example of how to format a question so that it will be answered. – davidlowryduda Jan 25 '12 at 1:54

Note that $x=d$ is a solution of your cubic equation (not the one you want, of course).