Hey it’s Professor Dave, let’s try some optimization problems.
One of the most common phrases you will hear in a mathematics classroom is “Why do I
need to know this?”
While I certainly can’t promise that every person watching this will use math every day
in their chosen careers, it is very easy to demonstrate the endless real-world applications
of calculus, which allow anyone to see exactly how powerful this discipline is when it comes
to solving complex problems.
And not just in physics.
This applies to business, too.
We may be trying to decipher a business strategy that will minimize manufacturing costs, or
maximize profits.
We may be trying to minimize a distance travelled or maximize an area of property.
In any of these cases, we should notice the words minimize and maximize.
These words mean that we are looking for a function’s local maximum or minimum, something
we already learned how to do.
Let’s look at an example and I’ll show you what I mean.
Let’s say that a farmer has twenty four hundred meters of fencing, and wants to use
this to enclose some rectangular plot of land by a river.
The farmer needs only three sides for the fence, as the plot can remain open to the
river itself, and they want the biggest plot possible.
What will be the dimensions for this plot that will maximize its area?
Well let’s take a few guesses just for fun.
Say we go one hundred meters up from the river on either side, leaving twenty two hundred
meters of fence to connect them.
Calculating the area of the rectangle, that’s two hundred twenty thousand square meters.
We can probably do better than this.
How about a square, with eight hundred meters per side?
That will give us six hundred forty thousand square meters.
Definitely an improvement.
But is this really the best we can do?
We could guess and check all day, but why don’t we just do some calculus instead?
Well first let’s assign some variables.
Let’s call these two sides x, and this side y.
We care about the area, and since this is a rectangle, area equals x times y.
We also know that these three sides must add up to twenty four hundred, because that’s
all the fencing we have to work with, so let’s say that two x plus y equals twenty four hundred.
Let’s solve for one of these variables, perhaps y.
Y must equal twenty four hundred minus two x.
Now we can plug this expression in for y in our area equation, and we get that the area
equals x times the quantity (twenty four hundred minus two x).
We distribute the x, and this is what we get.
Now we said that we want to get the biggest area possible.
That means we want the maximum possible value of this function.
Well we learned that to find a maximum or minimum, we take the first derivative and
see where it equals zero, so let’s do that.
The first derivative will be twenty four hundred minus four x.
We can just set this equal to zero, move things around a bit to solve for x, and we get x
equals six hundred.
Area will be at a maximum when x equals six hundred meters, so these two sides are six
hundred, making this side twelve hundred, and the maximum possible area of the rectangle
will be seven hundred twenty thousand.
So as you can see, taking a quick derivative and setting it equal to zero is much easier
than doing guess and check all day.
All we had to do was draw a diagram, identify the unknown, or what we are trying to solve
for, express this in terms of algebraic relationships, and then do calculus to find a maximum or
minimum value.
Let’s try another one.
Say a canned good manufacturer wants to mass produce a product in a cylindrical can that
must have a volume of 1.5 liters.
In order to keep costs as low as possible, they want to find the least amount of material
that can be used to achieve this particular volume.
What amount will this be?
Well this question is asking something about the surface area of the can, as that will
tell us how much material is needed to make one can.
How can we find the surface area of the cylinder?
Well we have two circular bases.
Those have an area of pi r squared, so putting them together, that’s two pi r squared.
Then we have this whole section, which is essentially a rectangle folded back on itself.
If we were to unfold this, it would have a base length of two pi r, as that’s the circumference
of the base, and a height which we can represent as h.
The area of the rectangle will simply be their product, two pi r times h.
The total surface area will therefore be the sum of these two, or two pi r squared plus
two pi r times h.
Now we would prefer to have everything in terms of r, so what can we do with h?
Well let’s recall that we know something about the volume of this cylinder.
That must be 1.5 liters, or fifteen hundred cubic centimeters.
How does volume relate to all this?
Well the volume of a cylinder is the area of the base times the height, and in this
case that’s pi r squared h, so let’s use this expression to solve for h in terms of
r. Dividing both sides by pi r squared, we get fifteen hundred over pi r squared.
So let’s plug that in for h in our surface area equation.
Then we can distribute this term, and we end up with two pi r squared plus three thousand
over r.
Now if this is the equation that gives us the surface area of the can in terms of its
radius, and we want to minimize this function, or find the smallest amount of material that
will yield this volume, all we have to do is take the derivative and set it equal to zero.
The derivative of the first term is four pi r, and then for this second term if we rewrite
this as three thousand r to the negative one, we get negative one times three thousand r
to the negative two, so we end up subtracting three thousand over r squared.
If we get a common denominator here, we can combine the terms, and we are left with this,
four pi r cubed minus three thousand, over r squared.
As we said, we want to find where this derivative equals zero, because that will give us the
lowest point on the function, where the surface area goes from decreasing to increasing.
So let’s set this derivative equal to zero, and solve for r.
If we multiply both sides by the denominator, it will just go away, since we are multiplying
by zero.
Then we have to add three thousand to both sides, divide by four to get seven hundred
fifty, then we can divide by pi, and lastly we take the cube root.
That leaves us with the cube root of seven hundred fifty over pi.
And although that may not look pretty, that’s our answer for the ideal radius of the can
that will keep costs at a minimum.
We can then plug this into another equation to solve for the height of the can as well,
and that will give us the ideal dimensions of the can.
The last thing we want to be aware of is that when the derivative of a function is zero,
there is a maximum or minimum at that point.
But we certainly must be able to know whether it is a maximum or minimum.
If we think we are finding the minimum cost of something but we are really finding the
maximum cost, our boss will be very angry with us.
So let’s learn something called the second derivative test.
Wherever a derivative is equal to zero, we can take the derivative again, to get the
second derivative.
If the second derivative is positive at that value, then that point is a local minimum,
because that part of the curve must be concave up.
If instead, the second derivative is negative at that value, the point is a local maximum,
because that part of the curve must be concave down.
If our answers make perfect sense without doing this, it may not be necessary, but if
in doubt, it doesn’t take much time to check the numbers in this manner.
So between related rates and optimization problems, we should now understand the power
of calculus, and the wide variety of its applications.
We will discover more things we can do with calculus later, for now let’s check comprehension.