MA001: College Algebra

With quadratics, we were able to algebraically find the maximum or minimum value of the function by finding the vertex. For general polynomials, finding these turning points is not possible without more advanced techniques from calculus. Even then, finding where extrema occur can still be algebraically challenging. For now, we will estimate the locations of turning points using technology to generate a graph.

Each turning point represents a local minimum or maximum. Sometimes, a turning point is the highest or lowest point on the entire graph. In these cases, we say that the turning point is a global maximum or a global minimum. These are also referred to as the absolute maximum and absolute minimum values of the function.

LOCAL AND GLOBAL EXTREMA

A local maximum or local minimum at $x=a$ (sometimes called the relative maximum or minimum, respectively) is the output at the highest or lowest point on the graph in an open interval around $x=a$. If a function has a local maximum at $a$, then $f(a) \geq f(x)$ for all $x$ in an open interval around $x=a$. If a function has a local minimum at $a$, then $f(a) \leq f(x)$ for all $x$ in an open interval around $x=a$.

A global maximum or global minimum is the output at the highest or lowest point of the function. If a function has a global maximum at $a$, then $f(a) \geq f(x)$ for all $x$. If a function has a global minimum at $a$, then $f(a) \leq f(x)$ for all $x$.

We can see the difference between local and global extrema in Figure 21.

Figure 21

Q&A

Do all polynomial functions have a global minimum or maximum?

No. Only polynomial functions of even degree have a global minimum or maximum. For example, $f(x)=x$ has neither a global maximum nor a global minimum.

EXAMPLE 11

Using Local Extrema to Solve Applications

An open-top box is to be constructed by cutting out squares from each corner of a $14$ cm by $20$ cm sheet of plastic and then folding up the sides. Find the size of squares that should be cut out to maximize the volume enclosed by the box.

Solution

We will start this problem by drawing a picture like that in Figure 22, labeling the width of the cut-out squares with a variable, $w$.

Figure 22

Notice that after a square is cut out from each end, it leaves a $(14−2w)$ cm by $(20−2w)$ cm rectangle for the base of the box, and the box will be w cm tall. This gives the volume

$\begin{aligned} V(w) &=(20-2 w)(14-2 w) w \\ &=280 w-68 w^{2}+4 w^{3} \end{aligned}$

Notice, since the factors are $w$, $20–2w$ and $14–2w$, the three zeros are $10$, $7$, and $0$, respectively. Because a height of $0$ cm is not reasonable, we consider the only the zeros $10$ and $7$. The shortest side is $14$ and we are cutting off two squares, so values $w$ may take on are greater than zero or less than $7$. This means we will restrict the domain of this function to $0 < w < 7$. Using technology to sketch the graph of $V(w)$ on this reasonable domain, we get a graph like that in Figure 23. We can use this graph to estimate the maximum value for the volume, restricted to values for $w$ that are reasonable for this problem – values from $0$ to $7$.

Figure 23

From this graph, we turn our focus to only the portion on the reasonable domain, $[0,7]$. We can estimate the maximum value to be around $340$ cubic cm, which occurs when the squares are about $2.75$ cm on each side. To improve this estimate, we could use advanced features of our technology, if available, or simply change our window to zoom in on our graph to produce Figure 24.

Figure 24

From this zoomed-in view, we can refine our estimate for the maximum volume to about $339$ cubic cm, when the squares measure approximately $2.7$ cm on each side.

TRY IT #6

Use technology to find the maximum and minimum values on the interval $[−1,4]$ of the function $f(x)=-0.2(x-2)^{3}(x+1)^{2}(x-4)$.