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Lets practice finding the absolute maximum and minimum with a word problem. Here we are
going to use the word problem that you were presented in the introduction to this lesson
so here's the scenario. To supplement your income as a college student you've decided
to start a business of driving people home from parties. If you cram them you can fit
20 people in your dad's old construction van. If your profit from driving X people is P
of X equals X times quantity 10 minus 0.5 X minus 20 how many people should you load
into the van to maximize your profit? You have some sort of profit formula. You are
making money from charging these people but there's also costs associated with driving
them around. You've got gas, you've got you know if they throw up you've got to clean
it up so anyway you've got this profit formula and you want to know what is the X value that's
going to be my absolute that's going to provide me with my absolute maximum profit. To answer
this question we are going to the steps we've developed to find the absolute maximum and
apply it to this function. First of all let me rewrite our expression down here. Profit
is X times the quantity 10 minus 0.5 X minus 20. Now lets think about what X values make
sense. When you are doing one of the problems from the book or one of the examples we've
seen so far you're given the interval on which to check for the absolute max or min but when
we have a story problem like this you want to think yourself about what X values make
sense. What X values make sense here? What interval of X values am I looking at? The
minimum number of people you could have is 0 right? You can't have negative people in
your van so 0 is going to be the lower bound on the interval and the problem says you could
fit 20 people maximum into the van so 20 is going to be the upper bound of the interval.
We want to look for the absolute maximum of P of X on the closed interval from 0 to 20.
How do we find the absolute maximum of a function? In general the absolute maximum or minimum
could happen at a critical point or it could happen at an end point so in the end we are
going to check all of the critical points and end points to see what the actual function
value is but we need to start by finding those critical points. Remember, critical points
are places where the derivative equals 0 or does not exist. Lets start by finding the
derivative of this function. Before I find the derivative I'm just going to distribute
this X. In this case its probably going to be a little easier to distribute than to use
the product rule so I'm going to have this 10 X minus 0.5 X squared minus 20. This is
just a quadratic equation. Pretty simple derivative in this case. P prime of X just equals 10
minus so I bring the 2 down that's going to give me one half times 2 which is just one
so this is just going to be 10 minus X for the derivative. Critical points. Let's first
think about whether there are any places where this derivative does not exist. This is just
10 minus X. That's just a line. Its continuous so it exists for all X value so there are
no critical points from that criteria. Next we want to think about the X values that make
the derivative 0 so I want to solve the equation 0 equal 10 minus X. Of course that's a pretty
easy equation to solve, we just get X equals 10 so X equals 10 is our critical point. I
should mention that as a reminder you want to always make sure that critical point fall
within the domain of your original function so in this case our original functions domain
is 0 to 20 and this does fall within that domain so we are good. X equals 10 is a critical
point. Now we have our critical points. The next step to finding the absolute extrema
then is to check the function value in this case the profit at the end points and the
critical points because those are the only possibilities for where that absolute maximum
could occur. I usually just make a chart to do this. We have our X values and then we
have our profit equation which was X times 10 minus 0.5 X minus 20 and the X values we
are checking are 10 0 and 20. Now I'm just plugging them in. I'm going to plug in 10
into this function. Of course you can always do this on a calculator but I always like
to practice doing things by hand. 10 times 10 minus one half times 10 is 5 and then we've
got this minus 20 so that's 10 times 5 minus 20 which is 50 minus 20 which is 30. If you
are driving 10 people you are making a profit of 30 dollars for that load of people. What
about 0? Here I've got 0 times 10 minus 0 minus 20 so that's 0 minus 20 or negative
20. If you are just driving around with zero people then you are losing 20 dollars. What
about if you cram full of 20 people? Lets see. Now we are plugging in 20 so 20 times
10 minus half of 20 is 10 minus 20 so here again we've got 0 minus 20 or negative 20.
If you cram 20 people in your van maybe they reek such havoc that it cost you 20 dollars
just to repair the damage. Now that we have our chart complete we simply inspect it and
find the maximum profit. In this case we are only looking for a maximum. Sometimes you
are looking for both a maximum and a minimum. Here we just want to find the maximum profit
so in this case it occurs at 10. Lets go back up and look at what our problem was asking.
Sometimes a problem will ask where the maximum occurs. Sometimes it will ask what the maximum
actually is. In this case we're asked how many people should you load into your van
to maximize your profit. In this case our answer would be 10 people so our answer then
is 10 people will maximize our profit. If we were asked how much that profit would be
then we would say it would be 30 dollars. This is how you would approach an absolute
extrema problem in general and specifically a word problem involving absolute extrema.