Locating and Classifying Local Extrema

Definition of Relative Max and Min

We now extend the definition of relative max and min to functions of two variables.

Definition 1.  A function f(x,y) has a relative maximum at (a,b) if there is a      small circle centered at (a,b) such that            f(a,b) > f(x,y) 2. There is a relative minimum at (x,y) if            f(a,b) < f(x,y)     for all (x,y) in the a small circle centered at (a,b).

For differentiable functions of one variable relative extrema occur at a point only if the first derivative is zero at that point.  A similar statement holds for functions of two variables, but the both partial derivatives must be 0.

Theorem   If f(x,y) has a relative maximum or minimum at a point P,       fx(P)  =  fy(P)  =  0

Example   

Let 

        f(x,y) = x² + xy - 2y + x - 1

then 
        f_x(x,y)  =   2x + y + 1        f_y(x,y)  =  x - 2

so that 

        x - 2  =  0

 or 

        x  =  2

Hence

        2(2) + y + 1 = 0  

so 

        y = 5

A possible extrema is (2,-5).

Notice that just as the vanishing of the first derivative does not guarantee a maximum or a minimum, the vanishing of the partial derivatives does not guarantee a relative extrema either.  But once again, the second derivative comes to the rescue.  We begin by defining a version of the second derivative for functions of two variables.

Definition We call the matrix:             the hessian and its determinant is        D = fxxfyy - fxy2

Now we are ready to state the second derivative test for functions of two variables.  This theorem helps us to classify the critical points as maximum, minimum, or neither.  In fact their is a special type of critical point that is neither a maximum nor a minimum, called a saddle point.  A surface with a saddle point is locally shaped like a saddle in that front and back go upwards from the critical point while left and right go downwards from the critical point.

Theorem (Second Derivative Test For Functions of Two Variables) If gradf = 0 we have the following D fxx Type > 0 > 0 Rel Min > 0 < 0 Rel Max < 0 any Saddle = 0 any Test Fails

Example:

Let 

        f(x,y) = -x² - 5y² + 8x - 10y - 13   

then 

        f_x  = -2x + 8        f_y  =  -10y - 10    

has zeros at     (4,-1)

we have 

        f_xx = -2,    f_yy = -10,    and    f_xy = 0 

Hence

        D = (-2)(-10) - 0 > 0

so f has a relative maximum at (4,-1)

Example:  Let

        f(x,y) = 4xy - x⁴ - y⁴

then 

        f_x = 4y - 4x³         f_y  =  4x - 4y³
 

We solve:

        4y - 4x³ = 0 

so that 

        y = x³  

Hence 

        4x - 4(x³)³ = 0

or

        x - x⁹ = 0 

so that 

        x = 1    or    x = 0    or    x = -1

plugging these back into 

        y = x³  

gives us the points 

        (1,1), (0,0) and (-1,-1)

We have 

        f_xx = -12x²,    f_yy = -12y²,    and    f_xy =  4

Hence 

        D = 144x²y² - 16

We write the table:

Point	D	fxx	Type
(1,1)	126	-12	Max
(0,0)	-16	0	Saddle
(-1,-1)	126	-12	Max

Minimizing Distance

Example

Find the minimum distance from the point (2,1,4) to the plane

        x + 2y + z = 5

Solution:  

We minimize the square of the distance:

        S = (x - 2)² + (y - 1)² + (5 - x - 2y - 4)²

We find the partial derivatives and set them equal to 0:

        f_x  =  2(x - 2) - 2(5 - x - 2y - 4)        f_y  =   2(y - 1) - 2(5 - x - 2y - 4)

so

        2x - 4 - 10 + 2x + 4y + 8 = 0    and    2y - 2 - 10 + 2x + 4y + 8 =  0

simplifying gives

        4x + 4y - 6 = 0    and    2x + 6y - 4 = 0
or
        2x + 2y - 3 = 0    and    2x + 6y - 4 = 0

subtracting the first from the second, we get:

        4y - 1 = 0 

so 

        y = 1/4

Hence 

        x = 7/4

Finally, we have

        s = (7/4 - 2)² + (1/4 - 1)² + (5 - 7/4- 2(1/4) - 4)² = 35/16 = 2.1875

Taking the square root gives that minimum distance of 1.479.
Note that by geometry, this must be the minimum distance, since the minimum distance is guaranteed to exist.
 

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