Identify the extrema of the function

Question

Identify any extrema of the function by recognizing its given form or its form after completing the square. Verify your results by using the partial derivatives to locate any critical points

and test for relative extrema.

g(x, y) = (x - 1)² + (y - 3)²

(x, y, z) =

A) relative minimum

B) relative maximum

C) saddle point

D) none of these

Answer 1

1) Find the critical points. Meaning, check the partial first derivatives with respect to each variable. If you can find coordinates both partial first derivates share that are equal to 0, you have found a critical point.

As an equation: g_x(a,b)=g_y(a,b)=0. Where (a,b) is the critical point and the subscript x and subscript y are the partial derivatives.

Work:

g_x=2(x-1)

2(x-1)=0

x=1

g_y=2(y-3)

2(y-3)=0

y=3

We know there is a critical point at (1,3) as that is the coordinates of where the partial derivates are equal to 0.

2) Find of what kind of point (a,b) is. We now use the following equation and set of rules to determine what the critical point (1,3) is.

D = D(a,b) = g_xx(a,b) x g_yy(a,b) - [g_xy(a,b)]²

1) If D>0 and g_xx(a,b)>0 there is a relative minimum at (a,b).

2) If D>0 and g_xx(a,b)<0 there is a relative maximum at (a,b).

3) If D<0 then (a,b) is a saddle point.

4) If D=0 then is may be any of those options, but we need to use other techniques to classify the critical point

To apply the equation:

D = D(a,b) = g_xx(a,b) x g_yy(a,b) - [g_xy(a,b)]²

g_x(1,3) = 2(x-1)

g_xx(1,3) = 2

g_y(1,3) = 2(y-3)

g_yy(1,3) = 2

g_x(1,3) = 2(x-1)

g_xy(1,3) = 0

Therefore,

D=2 x 2 - 0²

D=4

Since g_xx=2 which is greater than 0 and D=4 which is greater than 0, this falls into category 2, Relative Minimum.

The z-coordinate at (1,3) would be

z=g(x,y)=(x - 1)² + (y - 3)²

z=g(1,3)=(1-1)² + (3-3)²

z=0

Final Answer:

(1,3,0) is a relative minimum.