Uploaded by TheIntegralCALC on 19.07.2011

Transcript:

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!

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!