Let's use what we've learned about finding the minimums and
maximums of functions to solve some optimization problems.
And you might be wondering, Sal, what is an
optimization problem.
And my reply to you is well, the problems we're about to do
are optimization problems, and I think you'll understand
why they're called optimization problems.
Anyway so the first problem is find two numbers whose
product is a negative 16.
So the product is equal to minus 16.
And it says, and the sum of whose squares is a minimum.
So sum of squares is minimum.
So if you haven't figured out yet, these optimization
problems all impose some type of constraints, and then say,
for what numbers do we get a maximum value or minimum value.
So let's do this problem, and then we'll do a couple of more.
So two numbers whose product is negative 16 and the sum
of their squares is minimum.
So call those two numbers x and y.
So their product is minus 16, so we know that x times
y is equal to minus 16.
And then what's the sum of their squares.
Let's just say the some of squares, so I'll
just call that s.
The sum of their squares is going to be equal to x
squared plus y squared.
And so this is the function that we want to optimize.
And in this case we want to find the minimum value.
So how do we do this problem?
Well the first thing we want to do, we want to minimize this
function, but right now it's a function of two variables,
and we haven't done multivariable calculus yet.
So it would be easier if we could express this as a
function of one variable, and then we could optimize it with
respect that variable, and we could substitute backwards
et cetera, et cetera.
So how can we write the sum of the squares as a
function of one variable?
Well that's where this equation comes in handy.
We could substitute either for x or y.
Let's just substitute y, so we could rewrite this equation
dividing both sides by x as y is equal to minus 16 over x.
I just divided both sides of this by x.
If we take this and substitute it here, we get the sum of the
squares of the two numbers is going to be equal to x
squared plus y squared.
Well y is this, so what's this thing squared?
Minus 16 squared is 256 over x squared.
So that's the sum of the squares of the two numbers.
And so how do we learn to find minimums or maximums, or in
general optimize the function?
Well in general, a function hits a minimum or maximum
point, especially a local minimum or maximum point,
when it's derivative is 0.
So let's take the derivative of this and set it equal to 0, and
hopefully we find a minimum point.
And then we'll later show that it really is a minimum point.
So let's take the derivative of this with respect to x.
I'll just write s prime.
We could have written ds dx, but save on notation.
Save on my electronic ink.
So s prime is equal to-- so there's a slope of this
function at any time, at any x-- 2x plus-- this is the
same thing as 256 times x to the negative 2.
So what will be the derivative?
The derivative is going to be 256.
We multiply times exponent which is minus 2, and then we
decrement the exponent by 1 , so we get x to the minus 3.
So the derivative is equal to 2x plus-- no, not plus-- 2x
minus 512 times x to the minus 3.
And we want to find a local minimum or maximum, or
maybe a global one.
So we want to figure out where the slope is equal to 0, so
this is the slope, so let's set this equal to 0.
So we get 2x minus 512x to the minus 3 is equal to 0.
Let's see, we can manipulate that around a little bit.
We get 2x-- let me do it in a different color,
I'm tired of the yellow.
So you get going up here, we get-- just let's add 512 to
both sides-- so we get 2x is equal to 512x to the minus 3,
right, I just added 512 to both sides.
Now let's multiply both sides of this equation times x to the
third, just to get rid of this.
So if we multiply both sides of this equation by x to
the third, we get 2x to the fourth is equal to 512.
Right, x to the third times x to the minus
3, that just equals 1.
Divide both sides of this by 2, you get x to the
fourth is equal to 256.
And if it's not immediately clear what this is-- you know,
x is going to be the fourth root of 256-- if you don't know
that immediately, you can think, well, what's the
square root of 256.
Well it's 16.
And then what's the square root of that.
Well the square root of 16 is plus or minus.
So the square root of this is plus or minus 16, but then
a square root of at least a plus 16 is plus or minus 4.
So this x is equal to plus or minus 4.
If you don't believe me, take either minus 4 or positive
4 and take it to the fourth power and you will get 256.
So if x is plus or minus 4, what is y?
Well we just know that y is minus 16 divided by x,
so we could get x is 4.
If x is 4, then what is y?
It's minus 16 divided by 4, then y is equal to minus 4.
And of course if x was minus 4, then y is minus 16 divided by
minus 4, y would be equal to 4.
So it actually doesn't matter which one we take.
The two numbers are 4 and negative 4.
It doesn't matter what we call them, x or y.
The two numbers are 4 and minus 4.
And we are actually done the problem.
These are the two numbers where the sum of the squares-- well
actually, no we, are not done.
We know that we've reached some type of minimum or maximum
point here, and since they ask for minimum, and this was the
only point that we found where the slope was 0, we probably
solved the problem.
But let's prove to ourselves that we're really at a minimum
point on this function when x is equal to 4.
And how do we do that?
Well, we can take the second derivative of this function,
see what the second derivative is at x equals 4, and then see
if we're concave upwards or concave downwards.
This is the first derivative.
What is the second derivative?
s prime prime is equal to 2, and then minus 3 times minus
512, so that's plus, and what is that?
1,536 Right, plus 1,536 times x to the minus 4.
I actually you know should be writing s as a function of x.
I shouldn't be lazy with my notation.
These are all the function of x, but you get the point.
So what is s prime prime when x is equal to 4?
Well it's equal to 2 plus 1,536.
And then what's 4 to the minus 4 power?
We get 1/256.
And we don't even have to calculate this.
All we care about is this value positive or negative.
It And it's clearly positive.
We have a positive number plus a positive number divided
by a positive number.
So we have a positive number here.
So what do we know?
We know at the point x is equal to 4, this function here-- our
sum of squares function-- is concave upwards, and we know
that because the second derivative at x is equal
to 4 is positive.
And what does concave upwards look like?
Concave upwards looks like this.
And so x equals 4 is there, and so we are at a minimum point.
And so we know that the sum of squares are minimum.
And if you really want to make sure that you're not missing
something or we haven't done this wrong, just play around,
think of other numbers whose product is minus 16 and figure
out their sum of squares.
And I can pretty much guarantee you that it will be larger than
the sum of the squares of 4 of minus 4 and 4.
So actually, what are they?
What is s of 4?
What is the sum of squares of 4 and minus 4?
It's 4 squared is 16 plus 16 is 32.
That's the sum.
So let's try some other numbers.
What is the sum of 1 squared plus minus 16 squared.
Those two have the product minus 16, 1 and minus 16,
that is equal to what?
257.
Much higher number.
You could try negative 8 and 2, or negative 2 and 8, and
everything you'll find will have a higher sum of squares
than 4 and minus 4.
Anyway.
I was actually planning on doing three problems in this
video, but I'm reaching the 10 minute limit so I will finish
this video now, and I will do two more of these
optimization problems.
See you soon.