Absolute Maxima And Minima Problems

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Finding the absolute maxima and minima of functions (the biggest and smallest values a function attains) is important for many real world calculus applications. For example, companies want to maximize their profits, engineers want to minimize the drag on an airplane, computer scientists want to get your youtube video from the server to your computer as fast as possible. This type of mathematics can also address questions like why soap bubbles take the shapes they do, and these types of problems very quickly lead to the frontiers of mathematical physics. But enough of that, let’s look at some examples of finding the absolute max and min of functions.

Have a look at the function below, which is 3x^4-16x^3+18x^2 on the interval [-1,4].

Rendered by QuickLaTeX.com

Let’s consider the minima of this function first. It looks like this function has a local minimum at x=0 and an absolute minimum at x=3. Absolute here means that this is the smallest value the function attains on that interval. Local means that if you zoom way in it’s a minimum, but clearly the function gets smaller at x=3.

How could we do this if we didn’t have a graph of the function? If all we’re given is f(x)=3x^4-16x^3+18x^2 on the interval [-1,4], we need to find the critical points:

    \begin{align*} f'(x) &= 12x^3-3\cdot 16x^3+36x\\ & = 12x(x^2-4x+3)\\ & = 12x(x-1)(x-3) \end{align*}

Therefore f'(x) = 0 at x=0, x=1, and x=3. The first and last of these are the minima we saw above. It looks from the graph like x=1 is a maximum. Is that the absolute maximum of the function? No, clearly not! The function is bigger at x=-1 and x=4.

This is an important point: to find the absolute maximum and minimum of a function on a closed interval [a,b], first find the critical points of the function. Plug in these values to the function, and also plug in the endpoints x=a and x=b. The biggest number is the absolute maximum and the smallest number is the absolute minimum of the function.

It turns out that continuous functions on a closed, bounded interval always have an absolute maximum and an absolute minimum. However, that’s not true if the interval isn’t closed or isn’t bounded. For example, e^x has no maximum value on [0,\infty). It just keeps growing and growing. It’s also not hard to think of an example of a function that isn’t continuous which has no maximum or minimum value on an interval.

Here are some more problems to think about:

Q: What are the absolute maximum and minimum values of e^{-x} on the interval [-1, 1]? What about on the interval [0,\infty)?

Answer: if f(x)=e^{-x}, then f'(x) = -e^{-x}. This function is never zero, so it has no critical points. Since f is decreasing, on the interval [-1, 1] the absolute maximum is e and the absolute minimum is e^{-1}.

Q: Find the absolute maximum and minimum values of f(x)=1-x^2 when 0<x\leq 2. What about when -1\leq x\leq 2?

Answer: f'(x) =-2x here, so f has a critical point at 0. This isn’t in the set of x such that 0<x\leq 2, though. Since f'<0 when 0<x\leq 2, f is decreasing there. Since f has its maximum at 0, but this point isn’t in the domain, f has no absolute maximum on the domain. It does have an absolute minimum at f(2) = -3.

On the other hand, on the domain -1\leq x\leq 2, we need to check the critical point x=0 as well as the endpoints x=-1 and x=2. Computing, we get f(-1) = 0, f(0) = 1, and f(2) = -3. That means the absolute maximum of f is 1 and the absolute minimum is -3.