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Home Continuous Functions Maxima and Minima - Applications Examples Example 3: Inscribing a Cylinder Into a Sphere Sollution Two: Implicit Differentiation
 
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Sollution Two: Implicit Differentiation

Instead of eliminating h and expressing V as a function of x, we shall use the equations in their original form and find the critical points by implicit differentiation.

Step 1

V = πx2h, dV/dx = 2πxh + πx2 dh/dx.

We find dh/dx by implicit differentiation.

03_continuous_functions-229.gif , (h ≠ 0).

Then

03_continuous_functions-230.gif, (h ≠ 0).

Step 2

When h = 0 we have x = r, which is an endpoint. When h ≠ 0 we set dV/dx = 0 and solve for x.

2πxh - 4πx3/h = 0, xh - 2x3/h = 0,

xh2 = 2x3, x = 0 or x = ± h/√2.

We reject x = -h/√2 because x and h ≥ 0. x = 0 is an endpoint. Thus x = h/√2 is the unique interior critical point.

Step 3

We use the Direct Test.

At x = 0, V = 0.

At x = h/√2, V = πh3/2.

At x = r, V = 0.

CONCLUSION The maximum of V is at x = h/√2. At that point the ratio of x to h is x/h = 1/√2.

The second method of solution may be better in a problem where it is hard or impossible to find explicit equations for the dependent variables (like h and V) as functions of the independent variable.
Home Continuous Functions Maxima and Minima - Applications Examples Example 3: Inscribing a Cylinder Into a Sphere Sollution Two: Implicit Differentiation

Last Update: 2006-11-25