Introduction to Extreme Values

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Anything You Want to Talk About?

What do “dy” and “dx” mean when used in an equation such as “dy = f′(x) dx”?

There are 2 ways to think about what this means. One is that dy and dx are literally tiny little changes in x or y, so that when x changes by some small amount (i.e., dx), the corresponding change in y (i.e., dy) is the slope of f (i.e., f′(x)) times that change in x. Then “dy/dx” is literally a slope, i.e., rise over run.

The other way of thinking about differentials is a consequence of the first: if dy and dx are tiny changes in y or x, then they should be things you can do algebra on. In particular, you can take the fact that dy/dx is another way of representing f′(x), i.e., the two are equal, and so the unfamiliar “dy = f′(x) dx” is just what you get if you multiply both sides of the equation by dx:

SI

Today 5:00 - 6:30. The link will be announced around 4:30. The session will review material from today & Monday, then go over problem sets.

Extreme Values

From “Absolute Extrema” and “Local Extrema and Critical Points” in section 4.3 of our textbook, and this extreme values discussion.

The Main Idea

From the example in the discussion: Geneseo Widget Works is launching a new advertising campaign, and predicts that sales t days after the start of the campaign will be t²e^-t thousands of widgets per day. When should they expect sales to peak, and at what volume?

This is an extreme value problem because it involves finding where some function reaches its maximum or minimum — in this case, the value of t at which sales will reach a maximum.

Is the problem looking for a local or an absolute maximum? Absolute, because the context of the problem implies that you want the largest sales volume over all time, which is what an “absolute” extreme value is. A “local” extreme, in contrast, is just the largest or smallest value in some region, and there may be more extreme ones outside that region.

Procedure for solving extreme value problems for function f(x):

1. Find f′(x).
2. Solve f′(x) = 0 for x.
3. Identify places where f′(x) doesn’t exist.
4. Plug each x value from (2) and (3) into f; the extreme value is the largest or smallest (depending on what kind of extreme you want) of the results. Sometimes you also want the corresponding x values.

Apply that procedure to Geneseo Widget Works’ advertising campaign

Start by taking the derivative of the sales-by-time function. Notice in doing this that you use the chain rule to differentiate e^-t.

Then solve for places where the derivative equals 0. Our general strategy for this was to factor the derivative, keeping in mind that it will be 0 if any one of the factors is. So when we had simple enough factors we could solve for the t that would make that factor 0 without worrying about the other factors. But e^-t is never 0, so that factor doesn’t contribute any solution to the equation:

We should also check for any values of t at which the derivative is undefined, but in this case there are none.

Finally, plug each t value we found into the equation for sales, and identify the one that produces the largest number. In this case, since Geneseo Widget Works also wants to know the maximum sales volume, note the value of s(t) that goes with this t:

A Radio Example

Devices like cell phones or wifi cards that communicate over radio waves use something called a “tuned circuit” to select the radio frequency they “listen” to or “talk” on. Tuned circuits are based on 2 kinds of electrical component (for the electronically inclined, called “inductors” and “capacitors”) whose resistance to the flow of electricity is proportional to frequency (inductors) and proportional to the reciprocal of frequency (capacitors) of an electrical signal or radio wave. Having resistances that depend on frequency is the key to how tuned circuits can select one frequency in preference to others. In one simple sort of tuned circuit the overall resistance is the sum of these two effects, so I could model the overall resistance to electrical/radio signals in a minimal tuned circuit as R = f + 1/f, where f is frequency.

At what frequency does this circuit have its minimum resistance?

Start by finding the derivative of resistance. It’s easiest in doing this to treat the 1/f term as f^-1 and use the power rule to find the derivative.

At this point we ran out of class time, but see how much further you can get on this example through the extreme values discussion. We’ll pick up in class tomorrow wherever the discussion leaves off.

Next

More examples of critical points and extreme values.

No new reading, but do use the discussion to keep working on the radio tuning example and to raise any other points you want about extreme values.

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