Before we go into applied optimization, we want to begin by going through an example using an equation on a given interval. As we move forward, our goal with applied optimization will be to create model to solve which involves a problem of this type. Therefore, we will need to know how to solve these problems.

**Determine Absolute Extrema**

Here we will be finding the absolute maximum and minimum of the function

\begin{align*}

f(x)=\frac{x^{3}}{3}-x+2,

\end{align*}

on the interval \([0,3]\).

Before we start to find our solution, we first want to note that \(f(x)\) is a continuous function on the give interval \([0,3]\) and the interval \([0,3]\) is a closed an bounded interval. Therefore, we know that both the absolute maximum and absolute minimum of the function will exist. Furthermore, we know that these absolute extrema will occur at the local extrema which only occur at critical points. Therefore, we will have to find the critical points and compare them to see which gives the largest and smallest value.

**Critical Points**

We now know that the absolute extrema will have to exist and will occur at the critical points. We will, therefore, have to find the critical points. In order to do this, we recall that the critical values are values where the derivative is undefined or zero and the value is in the domain of the function. Hence, we need to find that the derivative is

\begin{align*}

f'(x)&=\left(\frac{x^{3}}{3}-x+3\right)’ \\

&=\left(\frac{x^{3}}{3}\right)’-(x)’+(3)’ \\

&=x^{2}-1.

\end{align*}

Now, we note that the \(f(x)\) is only defined on the interval \([0,3]\), since this is the domain given in the problem. Therefore, the function is undefined everywhere else. This means that the derivative does not exist at the endpoints \(x=0\) and \(x=3\). (Note that you can’t find secant lines from the left of 0 or the right of 3, so the two sided limit of slopes of secants does not exist). However, because \(x^{2}-1\) is a polynomial, this is defined everywhere else in the domain.

Now that we know where the derivative is undefined, we now need to find where the derivative is \(0\). We then find,

\begin{align*}

f'(x)&=0 \\

x^{2}-1&=0 \\

(x-1)(x+1)&=0 \\

x&= \pm 1.

\end{align*}

Before stating that these are critical points, we do need to make sure that these values are in the domain. In this case, we do have that \(1\) is in the interval \([0,3]\), but \(-1\) is not in \([0,3]\). Therefore, \(x=1\) is a critical value, but \(x=-1\) is not.

We have therefore found that the critical values are at \(x=0,1,3\).

**Comparing Values**

Now that we know all the critical values, we need to compare the function values at these points in order to determine the largest and smallest values. The best way to do this is to create a table, which we do below.

\(x\) | \(f(x)\) |
---|---|

\(0\) | \(3\) |

\(1\) | \(\frac{4}{3}\) |

\(3\) | \(9\) |

Because the largest \(y\)-value on table is \(9\), this tells us that the absolute maximum of the function on the given domain is \(9\). Furthermore, because the smallest \(y\)-value in the table is \(\frac{4}{3}\), this is the absolute minimum of the function. As a reminder, the maximum and minimum are the \(y\)-values not the \(x\)-values of the function.

As a quick check, we will graph the function and make sure that our answer make sense. Using the graph below, it appears that we have indeed found the correct absolute extrema.

**Conclusion**

I hope this helped you learn something and that you enjoyed the process. Since we now know how to find the absolute extrema of a function on a given interval, we can now work on solving applied optimization problems. Make sure to follow the blog, subscribe to the YouTube channel and share this with anyone else that would find this useful.