Global Extrema in Two Variables

Hi, everyone! Welcome back to integralcalc.com. Today we’re going to be talking about how
to find global extrema in a function of 2 variables. So we’ve got a function of the
two variables x and y and it’s equal to 2x^2 + 3y^2 – 12x – 6y + 9. And we’ve
been asked to find global extrema of this function if there are any. So what we need
to do first is take the first-order partial derivatives of the function f(x,y). So we’re
going to take the partial derivative of f with respect to x and we’re going to take
the partial derivative of f with respect to y. So I’m going to assume for the moment
that you roughly had to take partial derivatives and kind of move through it quickly.
But basically, when we take the partial derivative of f with respect to x, remember that we’re
treating x as the variable and holding y constant, treating it like a constant as if it were
a number like 3 or 4. So we’re only going to be looking at x as a variable. The derivative
of 2x^2 with respect to x is 4x. The derivative of 3y^2 will be zero because there’s no
x variable involved in that term. The derivative of -12x will be -12 and the derivative of
-6y + 9, both of those will be zero when taking the derivative because again, there’s no
x variable involved in those terms. So the partial derivative of f with respect to x
is 4x – 12. The partial derivative of f with respect to
y, in this case we’re treating y as the variable and holding x constant. The derivative
of 2x^2 will be zero because there’s no y variable involved. The derivative of 3y^2
will be 6y. The derivative of -12x will be zero. The derivative of -6y will be -6 and
the derivative of 9 will be zero because again, no y variable involved in that term.
So once we’ve taken the first-order partial derivatives of f, we need to set both of them
equal to zero and solve them each for their respective variables. So in this case, we’re
going to set 4x – 12 = 0 and solve for x. We’ll do so by adding 12 to both sides and
we’ll get 4x = 12. Then we’ll divide both sides by 4 to get x = 3. Now, we’re going
to set the partial derivative of f with respect to y equal to zero. We will add 6 to both
sides to get 6y = 6. And then divide both sides by 6 to get y = 1. Notice that we have
one answer for x and one answer for y. this tells us that those two things together make
up the coordinates of our extrema point. So we’ve got x = 3, y = 1. That means we have
an extrema point at (3,1). We don’t yet know if it’s a global maximum or a global
minimum or a sallow point but we know that we have an extrema point at (3,1).
So now that we’ve determined what the point is, we need to determine the quality of the
point whether it’s a max, a min or a sallow point. And in order to do that, we need to
take the second-order partial derivatives of f both the partial derivatives with respect
to x and y s well as the mixed second-order partial derivative. So let’s go ahead and
do that. Basically, what we’re going to do is look at the first-order partial derivative
of f with respect to x and take it with respect to x again. So the derivative of 4x – 12
with respect to x is simply 4. The second-order partial derivative with respect to y, again
we’ll be looking at the first-order partial derivative with respect to y and simply taking
the derivative of our answer with respect to y again. So the derivative of 6y – 6
with respect to y is just 6. And then we’re looking for the mixed second-order partial
derivative. Before, we took this one with respect to x and then x again, y and then
y again. This time, we need to either one of these two and then we take the derivative
with respect to the opposite variable. So we could either take the derivative of this
one with respect to y because we took the derivative with respect to x earlier. Or take
the derivative of this one with respect to x since we took the derivative with respect
to y earlier. So let’s go ahead and pick the first one. We’ll take the derivative
of 4x – 12 but this time with respect to y. That is just going to give us zero because
remember, when we’re taking the derivative with respect to y, we are treating y as the
variable and x as a constant. That means that both 4x and 12 are going to be constants.
The derivative of constants are zero. So this mixed second-order partial derivative is going
to be zero. So now that we’ve found the three second-order
partial derivatives, we can use these information to determine whether our extrema point (3,1)
is a max, a min, or a sallow point. In order to do that, we’re going to use this formula
here, D(x,y). And I’ve seen the formula written a bunch of different ways but basically,
the important thing to remember is that you’re going to multiply the second-order partial
derivative with respect to x by the second-order partial derivative with respect to y and then
subtract the mixed second order partial derivative squared. This squared exponent here is important
to remember. So basically, for us, that means that D(x,y) is going to be equal to the second-order
partial derivative with respect to x which we know to be 4 times the second order partial
derivative with respect to y which is 6 minus the mixed second-order partial derivative
which is 0 squared. Remember this squared exponent here. Don’t forget that. So 4(6)
– (0)^2 = 24. Basically, you take the first-order partial
derivatives, set them equal to zero, solve for the extrema point. Once you’ve got it,
take the second-order partial derivatives, all three of them then you plug your values
into this D(x,y) equation. So now you’ve got a value for D(x,y) and you can use this
and you can use the second-order partial derivative with respect to x to determine whether it’s
a max, a min or sallow point. So what you care about is whether or not those two values
are less than, more than or equal to zero. So look at D first always. You’re going
to look at D first and determine whether or not it is greater than or less than zero.
Since it’s greater than zero, that means that we have to look at the second-order partial
derivative with respect to x then I’ll go to the other possibilities in a second. Now,
we’re going to look at the second-order partial derivative with respect to x. it is
also greater than zero so when both of them are greater than zero, that means that our
extrema point (3,1) is going to be a global minimum. When D is greater than zero but the
second-order partial derivative of f with respect to x is less than zero, you’re looking
at a global maximum. And when D(x,y) is less than zero, you’re looking at a sallow point
and doesn’t matter what the value of the second-order partial derivative of f with
respect to x actually is. In this case, both are more than zero which means that the extrema
point (3,1) is going to be a global minimum. So that’s it. I hope this video helped you
guys and I will see you in the next one. Bye!