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Example 6

Find the maximum and minimum, if any, of the function

11_partial_differentiation-476.gif

Step 1

The domain is the whole (x, y) plane because the denominator is always positive.

Step 2

11_partial_differentiation-477.gif = -[2(x + y) + 2(x + l)][(x + y)2 + (x + 1)2 + y2] -2, 11_partial_differentiation-478.gif = -[2(x + y) + 2y][(x + y)2 + (x + 1)2 + y2] -2.

Step 3

The partial derivatives are zero when

2(x + y) + 2(x + 1) = 0, 2(x + y) + 2y = 0,

or

2x + y + 1 = 0, x + 2y = 0.

The critical point is

x = -⅔, y = ⅓, and z = 3.

Step 4

Let E be the hyperreal region -H ≤ x ≤ H, -H ≤ y ≤ H where H is positive infinite.

Step 5

At a boundary point of E where x = ±H, (x + 1)2 is infinite so z is infinitesimal. At a boundary point where y = ±H, y2 is infinite so again z is infinitesimal.

CONCLUSION

z has a maximum of 3 at the critical point (-⅔, ⅓). z has no minimum. The region E is sketched in Figure 11.7.12.

11_partial_differentiation-479.gif

Figure 11.7.12


Last Update: 2006-11-15