Maximum and Minimum Values: Examples

1. Find the and describe the critical point(s) for the following function: 2*x^2+y^2+8*x-6*y+20

Anwser:

First find the critical points, this is done by finding the x and y partial derivatives, and setting them equal to zero and solving as a system:

>	diff(2*x^2+y^2+8*x-6*y+20,x);

>	diff(2*x^2+y^2+8*x-6*y+20,y);

>	solve({4*x+8,2*y-6},{x,y});

We can see that this point is a mimimum by graphing. Due to the bounds, the point will be in the center of the graph

>	plot3d(2*x^2+y^2+8*x-6*y+20,x=-2..6,y=-1..7);

2.Find the extremma of the following function: -x^3+4*x*y-2*y^2+1

Anwser:

First, find the critical points in the same way as above

>	diff(-x^3+4*x*y-2*y^2+1,x);

>	diff(-x^3+4*x*y-2*y^2+1,y);

>	solve({-3*x^2+4*y,4*x-4*y},{x,y});

Now we use the second derivative test. We start by finding all of the second derivatives:

>	diff(-x^3+4*x*y-2*y^2+1,x,x);

>	diff(-x^3+4*x*y-2*y^2+1,x,y);

>	diff(-x^3+4*x*y-2*y^2+1,y,y);

Now using the second derivative test formula, we check each of the points:

>	eval((-6*x)*(-4)-(4)^2,x=0);

Thus, (0,0) is a saddle point.

>	eval((-6*x)*(-4)-(4)^2,x=4/3);

Because 16 is greater than zero, and fxx is negitive, it is a maximum. This can be seen a in agraph of the function:

>	plot3d(-x^3+4*x*y-2*y^2+1,x=0..3,y=0..3);