In Example 2.4 we considered the health status function y = f (x1, x2) = −x 2

1 + 2x1 − x 2

2 + 4x2 + 5. (2.99)

The first-order conditions for a maximum are f1 = −2x1 + 2 = 0, f2 = −2x2 + 4 = 0

(2.100)

or x

*1

= 1, x

*

2 = 2.

(2.101)

The second-order partial derivatives for Equation 2.99 are f11 = −2, f22 = −2, f12 = 0.

(2.102)

These derivatives clearly obey Equations 2.97 and 2.98, so both necessary and sufficient conditions for a local maximum are satisfied.16 QUerY: Describe the concave shape of the health status function, and indicate why it has only a single global maximum value.