Question

Discuss the maximum or minimum values of u, when u=X3+Y3-3aXY

Answer 1

Critical Points are 2 in Number

Step-by-step explanation:

First, we find the critical points, that is points where $\Delta f$ is either $\overrightarrow{0}$ or undefined. For this, we compute the partials of f.

$\frac {\partial f}{\partial x}(x,y) = 3x^2 - 3y$

$\frac {\partial f}{\partial x}(x,y) = 3y^2 - 3x$

The partials are always defined. They are 0 when

$x^2 - x =0$

$y^2 - x = 0$

Solving for y in the first equation and replacing the second equation gives

$x^4 - x = 0$

$x(x^3-1) = 0$

x = 0 or x = 1

Since $y= x^2$ (first equation), we see that when x=0, y=0 and when x=1, y=1.  

So there are two critical points. They are (0,0) and (1,1).  

• To test these critical points, we need to compute the second partials of f.

$\frac {\partial^2 f}{\partial^2 x}(x,y) = 6x$

$\frac {\partial^2 f}{\partial^2 y}(x,y) = 6y$

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

We evaluate these partials at the critical points. When there are several points to consider, this is best done using a table such as the one below.

• So, we see that f(1,1) = -1 is a local minimum. f(0,0) = 0 is a saddle point, it is not a local extreme value.