Skip to main content

Section 7.3 Global Extrema

Definition 7.3.1. Global Extrema.

Consider the function of two variables \(f(x,y)\) on the domain D.

  • If there exists some point \((x_0,y_0)\) in \(D\) such that \(f(x,y) \leq f(x_0,y_0)\) for all points \((x,y)\) in \(D\) then the function has a global maximum at \((x_0,y_0)\text{.}\)

  • Similarly, if there exists some point \((x_0,y_0)\) in \(D\) such that \(f(x,y) \geq f(x_0,y_0)\) for all points \((x,y)\) in \(D\) then the function has a global minimum at \((x_0,y_0)\text{.}\)

The graph of the function

\begin{equation*} f(x,y) = x(5x+2)(3y-2)e^{-\left(\frac{x^2+y^2}{5} \right)} \end{equation*}

over the domain \(D = \left \{ (x,y): -5 \leq x \leq 5, \, -5 \leq y \leq 5 \right \}\) is shown below.

Figure 7.3.3.

Before discussing global extrema for functions of two variables, recall the situation for a function of one variable \(y=f(x)\text{.}\) If \(f(x)\) is continuous on the closed interval \(I=[a,b]\) then \(f(x)\) is guaranteed to have both a global maximum and a global minimum on \(I\text{.}\)

Figure 7.3.4.

These global extrema can be found by evaluating \(f(x)\) at

  • All of the critical points of \(f(x)\) in \(I\text{,}\) and

  • The endpoints of \(L\text{.}\)

The procedure for finding the global extrema of functions of two variables is very similar and is based on the following theorem.

Note that a closed region, \(D \subset \mathbb{R}^2\text{,}\) is a region in the plane that contains its boundary. For example in the diagram below Region \(D_1\) would be a closed region whereas Region \(D_2\) is not closed.

Figure 7.3.6.

A bounded region, \(D \subset \mathbb{R}^2\text{,}\) is a region in the plane that doesn’t extend to infinity in any direction. For example in the diagram below Region \(D_1\) would be a bounded region whereas Region \(D_2\) is not bounded.

Figure 7.3.7.

Remark 7.3.8. Locating Global Extrema.

To locate the global extrema of the continuous function \(f(x,y)\) on the closed and bounded domain \(D \subset \mathbb{R}^2\text{:}\)

  • Find all of the critical points in the interior of \(D\text{;}\)

  • Find the maximum and minimum values of \(f(x,y)\) on the boundary of \(D\text{;}\)

  • Evaluate \(f(x,y)\) at each of the above points and compare.

Find the global extrema of the function \(z(x,y) = x^2 + 2xy + 3y^2\) on the closed triangular region \(D\) with vertices \((-1,1)\text{,}\) \((2,1)\) and \((-1,-2)\text{.}\)

Answer.

Global maximum: \(17\) at \((x,y) = (-1,2)\)

Global minimum: \(0\) at \((x,y)=(0,0)\text{.}\)

Solution.
Figure 7.3.10.

Firstly note that \(D\) is a closed and bounded region in the plane and so we can use the method outlined above. So begin by finding the critical points of \(z\text{.}\) Here

\begin{equation*} z_x = 2x+2y \: \text{ and } \: z_y = 2x+6y\text{.} \end{equation*}

Critical points occur when \(z_x = z_y = 0\) and so this function has only one critical point at \((x,y) = (0,0)\text{.}\) This is inside \(D\) and so we evaluate the function at this point, i.e.

\begin{equation*} z(0,0) = 0\text{.} \end{equation*}

To find the maximum and minimum values of the function on the boundary we will have to consider the 3 sides of the triangle separately. Firstly, consider the side of the triangle defined by

\begin{equation*} y=1, \: -1 \leq x \leq 2\text{.} \end{equation*}

On this interval we think of \(z\) as a function of \(x\) only, i.e.

\begin{equation*} z(x) = x^2 + 2x + 3\text{.} \end{equation*}
Figure 7.3.11.

This has a maximum value \(11\) at \(x=2\) and minimum value \(2\) at \(x=-1\text{.}\)

Next consider the side of the triangle defined by

\begin{equation*} x=-1, \: -2 \leq y \leq 1\text{.} \end{equation*}

On this interval we think of \(z\) as a function of \(y\) only, i.e.

\begin{equation*} z(y) = 1-2y+3y^2\text{.} \end{equation*}
Figure 7.3.12.

Again, using the technique given above for locating the global extrema for a function of one variable (or by looking at the graph) we find that the largest value of \(z\) occurs at \(y=-2\) (giving \(z=17\)) and the smallest value of \(z\) occurs at \(y=1/3\text{,}\) (giving \(z=2/3\)).

Finally on the interval defined by

\begin{equation*} y=x-1, \: -1 \leq x \leq 2\text{,} \end{equation*}

we can think of \(z\) as a function of \(x\) only, i.e.

\begin{equation*} z(x) = 6x^2-8x+3\text{.} \end{equation*}
Figure 7.3.13.

For this function the global maximum is \(17\) at \(x=-1\) and the global minimum is \(1/3\) at \(x=2/3\text{.}\)

On comparing the value of the function \(z(x,y) = x^2 + 2xy + 3y^2\) at each of the global extrema on the sides of the triangle and at the critical point inside the region we conclude that the function has a global maximum of \(17\) at \((x,y) = (-1,2)\) and a global minimum of \(0\) at \((x,y)=(0,0)\text{.}\)

Exercises Example Tasks

1.

Find the global extrema of \(z=2x^2 + x +y^2 - 2\) on \(D = \left \{ (x,y): x^2+y^2 \leq 4 \right \}\text{.}\)

2.

Find the global extrema of \(R(x,y) = x \sqrt{8-x^2-y^2}\) on \(D = \left \{ (x,y): x^2+y^2 \leq 8 \right \}\text{.}\)