Question 11.6: (a) Find the extrema of the function z = - x² - y². (b) Find...

(a) To find the extrema we must use Equations 11.63. Since ∂z/∂x = – 2x and ∂z/∂y = – 2y, it follows that ∂z/∂x = ∂z/∂y = 0 at x = y = 0, at which point z = 0. Since z is negative everywhere else (see Figure 11.9 ), it is clear that

\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0                (11.63)

the extreme value, z = 0, is the maximum value

(b) The constraint may be written g = y – 2x – 3. Clearly, g = 0. Multiply the constraint by the Lagrange multiplier η and add the result (zero!) to the function – (x² + y²) to obtain

h = – (x² + y²) + η(y – 2x – 3)

This is a function of the three variables x, y and η. For it to be stationary, the partial derivatives with respect to all three of these variables must vanish. First we have

Setting this equal to zero yields

x = -η            (a)

Next,

For this to be zero means

y=\frac{\eta}{2}               (b)

Finally

Setting this equal to zero gives us back the constraint condition,

y = – 2x – 3 = 0                 (c)

Substituting (a) and (b) into (c) yields η = 1.2, from which (a) and (b) imply,

x = – 1.2           y = 0.6            (d)

These are the coordinates of the point on the line y = 2x + 3 at which z = – x² – y² is stationary. Using (d), we find that

z = – 1.8

at this point.
Figure 11.9 is an illustration of this problem, and it shows that the computed extremum (a maximum, in the sense that small negative numbers exceed large negative numbers) is where the surface z = – x² – y² is closest to the line y = 2x + 3, as measured in the z-direction. Note that in this case, Equation 11.67 yields d²h = – 2dx² – 2dy², which is negative, confirming our conclusion that the extremum is a maximum.

d^{2} h=\frac{\partial^{2} h}{\partial x^{2}} d x^{2}+2 \frac{\partial^{2} h}{\partial x \partial y} d x d y+\frac{\partial^{2} h}{\partial y^{2}} d y^{2}                (11.67)