Maxima and Minima

The Essentials

To find local minima and maxima and saddle points, the critical points of the function are identified. Critical points occur either when both partial derivatives are 0 or when either partial derivative doesn't exist. The second derivative test can be used on each of the critical points:

\[ D=f_{xx}f_{yy}-(f_{xy})^2 \]

If \( D > 0 \) and \( f_{xx} > 0 \), the critical point is a minimum.

If \( D > 0 \) and \( f_{xx} < 0 \), the critical point is a maximum.

If \( D < 0 \), the critical point is a saddle point.

If \( D=0 \), the test is inconclusive.

The absolute minimum and maximum of a function over a certain domain will be on a critical point or on the boundary of the domain

Example

Find the absolute minimum and maximum of the function over the boundary of the rectangle: \( -1\leq x\leq1 \) and \( 0\leq y\leq3 \):

\[ f(x,y)=x^2-2xy+2y^2+4x-8y+4 \]

First, we find the critical points and their values:

\[ f_x=2x-2y+4f_y=-2x+4y-8 \]
\[ \begin{cases} 2x-2y+4=0 \\ -2x+4y-8=0 \\ \end{cases} \]
\[ \begin{cases} x=0 \\ y=2 \\ \end{cases} \]

This is the only critical point inside since the function exists for the whole domain. To find the points on the boundary we can parameterize each of the four lines. Then we set the derivative, that is with respect to \( t \), equal to 0. The first is \( x(t)=t \) and \( y(t)=0 \) from \( -1\leq t\leq 1 \), which gives us:

\[ t^2+4t+4=0 \]
\[ 2t+4=0 \]
\[ t=-2 \]

Which isn't between -1 and 1 and so is not on the line segment. The second line segment \( x(t)=t, y(t)=3 \) from \( -1\leq t\leq 1 \):

\[ t^2-2t-2=0 \]
\[ 2t-2=0 \]
\[ t=1 \]

The point corresponding to \( t=1 \) is on the domain and so can be an absolute minimum or maximum. The corresponding point is \( (1,3) \). The third line segment \( x(t)=-1, y(t)=t \) from \( 0\leq t\leq 3 \):

\[ 2t^2-6t+1=0 \]
\[ 4t-6=0 \]
\[ t=\frac{3}{2} \]

The point corresponding to \( t=\frac{3}{2} \) is between 0 and 3 and so can be a minimum or maximum. The corresponding point is \( (-1,\frac{3}{2}) \). The fourth line segment \( x(t)=1, y(t)=t \) from \( 0\leq t\leq3 \):

\[ 2t^2-10t+9=0 \]
\[ 4t-10=0 \]
\[ t=\frac{5}{2} \]

So the point \( (1,\frac{5}{2}) \) could also be a minimum or maximum. In addition to these points, we have to check the four corner points as well.

\[ \begin{array}{ccc} f(0,2)&=&-4 \\ f(-1,\frac{3}{2})&=&-\frac{7}{2} \\ f(1,\frac{5}{2})&=&-\frac{7}{2} \\ f(-1,0)&=&1 \\ f(-1,3)&=&1 \\ f(1,3)&=&-3 \\ f(1,0)&=&9 \\ \end{array} \]

The point at (0, 2) is the absolute minimum, and the point at (1, 0) is the absolute maximum

Practice

Find the absolute extrema for the function \( f(x,y)=x^2+y^2-4x \) on the domain \( x^2+y^2\leq 9 \).

Solutions:

Absolute minimum: \( f(2,0)=-4 \)

Absolute maximum: \( f(0,3)=9 \)

More Resources

Openstax: Maixima/Minima Problems External Link Icon