To find the extreme values of a function such as f (x, y) = x3 -3xy + y3, it is sufficient to locate all pairs (x, y) where the partial derivatives are zero.
Evaluate
f (x, y) = 3x2 -3y
f (x, y) = -3x + 3y2
Enter the equations 3x2 - 3y = 0 and -3x + 3y2 = 0 into a display or a ×1 matrix, then from the Solve submenu, choose Exact.
Solve + Exact
3x2 - 3y | = | 0 | |
-3x + 3y2 | = | 0 |
Solution is :x = 0, y = 0
,
x = 1, y = 1
,
x = - ρ -1, y = ρ
where ρ is a root of Z2 + Z + 1
Note that
Dxxf (x, y)Dyyf (x, y) - Dxyf (x, y)
= 36xy - 9; hence (0, 0) represents a saddle point because - 9 < 0; and (1, 1) represents a local minimum because 36 - 9 > 0, and
Dxxf (x, y) = 6x > 0. The roots of Z2 + Z + 1 are
Z = -
+
i
and
Z = -
-
i
so there are no other real extreme values.
You can visualize the local minimum at (1, 1) by generating a plot of the surface. To get the following plot, with the insertion point in the expression x3 -3xy + y3, click itbpF0.3009in0.3009in0.0701in3dplot.wmf; in the Plot Components page of the Plot Properties tabbed dialog set the Domain Intervals to -1≤x≤2 and -1≤y≤2; choose Patch and Contour.
Plot 3D + Rectangular
x3 -3xy + y3
dtbpF3in2.0003in0pt
The level curve x3 -3xy + y3 = 0 goes through the point (0, 0, 0). To get a better view of this level curve, do the following.
Plot 2D + Implicit
x3 -3xy + y3 = 0
dtbpF3in2.0003in0pt
To get a better idea of where the z-values are positive and where they are negative, enter the equation x3 -3xy + y3 = - .5 and drag it to the frame. Note that the z-values on the surface z = x3 -3xy + y3 are negative inside the loop in the first quadrant and in the lower left corner of the xy-plane.