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Solution One: Eliminating One Variable

Express V as a function of x by eliminating h.

03_continuous_functions-218.gif

The problem is to find the maximum of V in the interval 0 ≤ x ≤ r.

Step 1

03_continuous_functions-219.gif , (x < r).

Step 2

There is one critical point at x = r, where dV/dx does not exist. We set dV/dx = 0 and solve for x to find the other critical points.

03_continuous_functions-220.gif , 4πx(r2 - x2) - 2πx3 = 0,

2πx(2r2 - 3x2) = 0, x = 0 or 03_continuous_functions-221.gif

We reject 03_continuous_functions-223.gif because 0 ≤ x ≤ r. The only interior critical point is 03_continuous_functions-222.gif

Step 3

We use the Direct Test.

At x = 0, V = 0.

At x =03_continuous_functions-224.gif, 03_continuous_functions-225.gif

At x = r, V=0.

CONCLUSION

The maximum of V is at 03_continuous_functions-226.gif (see Figure 3.6.5). At that point,

03_continuous_functions-227.gif

Then the ratio of x to h is

x/h = 1/√2.

Notice that, as we expected, this number does not depend on r.

03_continuous_functions-228.gif

Figure 3.6.5


Last Update: 2006-11-25