Extreme Value on a Surface

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.

$\blacktriangleright$ Evaluate

${\frac{{\partial }}{{\partial x}}}$f (x, y) = 3x2 -3y


${\frac{{\partial }}{{\partial y}}}$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.

$\blacktriangleright$ Solve + Exact


3x2 - 3y = 0  
-3x + 3y2 = 0  

Solution is : $\left\{\vphantom{ x=0,y=0}\right.$x = 0, y = 0$\left.\vphantom{ x=0,y=0}\right\}$,$\left\{\vphantom{ x=1,y=1}\right.$x = 1, y = 1$\left.\vphantom{ x=1,y=1}\right\}$,$\left\{\vphantom{ x=-\rho -1,y=\rho }\right.$x = - ρ -1, y = ρ$\left.\vphantom{ x=-\rho -1,y=\rho }\right\}$ where ρ is a root of Z2 + Z + 1

Note that Dxxf (x, y)Dyyf (x, y) - $\left(\vphantom{ D_{xy}f(x,y)}\right.$Dxyf (x, y)$\left.\vphantom{ D_{xy}f(x,y)}\right)^{{2}}_{}$ = 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 = - ${\frac{{1}}{{2}}}$ + ${\frac{{1}}{{2}}}$i$\sqrt{{3}}$ and Z = - ${\frac{{1}}{{2}}}$ - ${\frac{{1}}{{2}}}$i$\sqrt{{3}}$ 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.

$\blacktriangleright$ 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.

$\blacktriangleright$ 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.



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