Computing Extreme Values over a region

If $f\left(x,y\right)$ is continuous in some closed, bounded set $D$ in $R^2$ then there are points in $D$, $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ so that $f\left(x_1,y_1\right)$ is the absolute maximum and $f\left(x_2,y_2\right)$ is the absolute minimum of the function in $D$

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3 Step Way to Find Extreme Values

1. Find all the critical points of the function that lie in the region $D$ and determine the function value at each of these points.
   
   
2. Find all extrema of the function on the boundary. This usually involves the Calculus I approach for this work.
   
   
3. The largest and smallest values found in the first two steps are the absolute minimum and the absolute maximum of the function.

Example

Find the absolute minimum and absolute maximum of $f\left(x,y\right)=2x^2-y^2+6y$ on the disk of radius 4, $x^2+y^2\le 16$

First note that a disk of radius 4 is given by the inequality in the problem statement. The “less than” inequality is included to get the interior of the disk and the equal sign is included to get the boundary. Of course, this also means that the boundary of the disk is a circle of radius 4.

Let’s first find the critical points of the function that lies inside the disk. This will require the following two first order partial derivatives.

$$f_x=4x{f}_y=-2y+6$$

To find the critical points we’ll need to solve the following system.

$$\begin{array}{cc}4x& =0\\ -2y+6& =0\end{array}$$

This is actually a fairly simple system to solve however. The first equation tells us that $x=0$ and the second tells us that $y=3$. So, the only critical point for this function is $\left(0,3\right)$ and this is inside the disk of radius 4. The function value at this critical point is,

$$f\left(0,3\right)=9$$$$\begin{array}{rlr}y=-4:& x^2& =16-16=0\Rightarrow x=0\\ & y=4:& x^2& =16-16=0\Rightarrow x=0\\ & y=1:& x^2& =16-1=15\Rightarrow x=\pm \sqrt{15}=\pm 3.87\end{array}$$

The function values for $g\left(y\right)$ then correspond to the following function values for $f\left(x,y\right)$.

$$\begin{array}{cccc}g\left(-4\right)& =-40& \Rightarrow & f\left(0,-4\right)=-40\\ g\left(4\right)& =8& \Rightarrow & f\left(0,4\right)=8\\ g\left(1\right)& =35& \Rightarrow & f\left(-\sqrt{15},1\right)=35\text{and}f\left(\sqrt{15},1\right)=35\end{array}$$

Note that the third one actually corresponded to two different values for $f\left(x,y\right)$ since that $y$ also produced two different values of $x$.

So, comparing these values to the value of the function at the critical point of $f\left(x,y\right)$ that we found earlier we can see that the absolute minimum occurs at $\left(0,-4\right)$ while the absolute maximum occurs twice at $\left(-\sqrt{15},1\right)$ and $\left(\sqrt{15},1\right)$.