What is this question asking? Critical points of multi variable functions?

Ans. if y=f(x) is differentiable in a [a,b], then y'(dy/dx)=0 is to find the value of

x=x1 for the critical point which is either a maximum y(x1) or a minimum y(x1).

y"(d^2y/dy^2)<0 at x=x1=>y(x1) is a miximum.

y">0 at x=x1=> y(x1) is a minimum.

y"=0 at x=x1=> y(x1) is a point of inflection.

So, not all derivatives are 0.

e.g. Let y=x^4-6x^2+6

y'=4x^3-12x=0=>

4x(x^2-3)=0=>

x=0 or x=+/-sqr(3)

y"=12x^2-12=12(x^2-1)

At x=0, y"(0)=-12<0=>

y(0)=6 is a max.

At x=+/-sqr(3), y"[+/-sqr(3)]=12(2)=24>0=>

y[+/-sqr(3)] are both a min.

In this example, there is no point of

inflection, because no x from y'=0 makes

y"=0.

The Weierstrass theorem assures us of the existence of the global minimum and the global maximum.

These points are to be found between:

i) singular points, ie where the gradient exists. In our case = Ø

ii) stationary points, ie where the gradient is canceled. In our case in O (0,0)

iii) at the border. In our case the border is composed of three segments (sides of the triangle T)

If we express the segments through the line equation on which it lies, we reduce the function into a single variable.

We know, again for Weirestrass, that there is a minimum and a maximum on each segment. Figure them out.

Thus we have an L list of a finite number of points.

The exercise ends with the L list.