Hi, everyone!.
Welcome back to integralcalc.com.
Today, we’re going to be talking about how to use the second derivative test to find
local maxima and minima.
And in this particular problem, we’ve been given the function f of x,y equals x squared
minus xy plus y squared plus 3y minus 1.
And we’re going to use second derivative test again to find whether or not this function
has any local maxima or minima.
So to do that, the first thing we need to do is take the first order partial derivatives
with respect to both x and y.
Because this function has two variables, we have to take partial derivatives.
So we’ll take the partial derivative with respect to x first and then with respect to
y.
Remember that when we take the partial derivative with respect to x, we’re going to be treating
x as the variable and holding y as a constant.
So we’ll take the derivative term by term.
The derivative of x squared with respect to x is regular.
We don’t have to worry about it because there’s no y involved in that term so it’s
just 2x.
When we get to negative xy here, remember that x is the variable.
If we take the derivative with respect to x, and if it were just x, the derivative of
course would be 1.
So that part’s going to become 1.
y is going to remain as a coefficient so that negative xy becomes negative y.
Then y squared, 3y and negative 1, the derivative of those with respect to x will all be zero
because there’s no x variable involved in any of those three terms.
So the partial derivative with respect to x is just 2x minus y.
When we take the partial derivative of f with respect to y, remember that we’ll be holding
y as the variable and holding x as a constant.
So of course, the derivative of x squared will be zero because there’s no y variable
involved in that term.
The derivative of negative xy with respect to y will be negative x.
Again, the derivative of y would be 1.
The negative x remains as a coefficient on that y variable so we just have negative x
times 1 which of course just gives us negative x.
And then, the rest of the terms are easy because there’s no x variables involved so we have
plus 2y plus 3 and of course negative 1, the derivative of that is zero because it’s
a constant.
So now that we have our first order partial derivatives, we need to go ahead and solve
these system of equations for a coordinate at which we can evaluate the function to see
whether or not it’s a maximum or a minimum.
The way we’re going to do that is by both setting these partial derivatives equal to
zero.
And now, we’re going to solve them as a system of equations.
You remember from algebra, simultaneous equations, so what I’m going to do is I’m going to
move all constants to the right-hand side which is going to get me 2x minus y equals
zero and then for the second equation here, we’ll have negative x plus 2y and I’m
going to subtract 3 from both sides to get negative 3.
Now what I want to do is multiply the second function by 2 and there’s a lot of ways
you could do this but I’m going to multiply the second function by two so that we end
up with negative 2x plus 4y is equal to negative 6.
And the reason that I chose to do that is because now I have a positive two x and a
negative 2x which means that if I add these equations together, then I’ll get 2x plus
a negative 2x which will give me a zero.
In other words, my x’s are going to cancel and I’ll just be left with y’s so I can
solve for the y variable, plug that back in and solve for x.
So when I end up with negative y plus 4y, I’ll get plus 3y and then over here on the
right hand side, zero plus a negative 6 gives me negative 6.
When I divide both sides by 3, I get y equals negative 2.
So I’ve got y equals -2.
I now need to solve for my x variable.
So I need to go ahead and plug y equals -2 back into any of these equations.
I’ll plug into these one here because it’s pretty simple.
So I’m going to say, 2x minus a negative 2 is going to give me zero which is going
to give me 2x plus 2 equals zero.
I’ll subtract 2 from both sides to get 2x equals negative 2 and then divide both sides
by 2 to get x equals negative 1.
So now that I’ve solved for my two variables here, I know that I’m going to be evaluating
at the point (-1,-2), where x is equal to negative 1 and y is equal to negative 2.
So that point represents probably, either a local maximum, a local minimum, a sallow
point, or if the second derivative test is inconclusive, I won’t be able to tell what
that point is if anything.
But that’s the point that I’m going to be looking at.
And sometimes, with the second derivative test, when you solve the system of equations
here, you may end up with multiple points in which case, you’ll need to evaluate each
of them.
So in this case, we have 1.
What we need to do next is take the second order partial derivatives.
So I’m going to take the second order partial derivative with respect to x which is going
to look like this.
The second order derivative with respect to y and then I’m also going to take the mixed
second order partial derivative.
So remember, when I’m taking the second order partial derivative with respect to x,
I have the first order partial derivative with respect to x up here.
So I’m going to take the derivative of that with respect to x again.
When I take the derivative with respect to x of 2x minus y, I’ll just get 2.
The derivative of 2x would be 2 and he derivative of would be zero.
So that is the second order partial derivative with respect to x.
Then for the second order partial derivative with respect to y, I’ve got the first order
partial derivative with respect to y and I’m going to take the derivative of that with
respect to y again.
The derivative of negative x with respect to y would be zero.
The derivative of 2y would be 2 and the derivative of 3 would be zero because of course, it’s
a constant.
So that’s the second order partial derivative with respect to y.
And now I need to take the mixed second order partial derivative and the way I’m going
to do that is take either one of my first derivatives and then take the derivative of
that with respect to the opposite variable of what I took the derivative of it before.
So in other words, I can take the derivative of the first order partial derivative with
respect to x, take it with respect to y or I can take the derivative of the first order
partial derivative with respect to y but this time with respect to x.
Either way, I should get the exact same answer.
So I’ll go ahead and take the derivative of the first order partial derivative with
respect to x but I’m going to take it with respect to y.
So the derivative of 2x minus y with respect to y is 2x will become zero because there’s
no y variable involved, negative y will become negative 1 and that’s going to be my mixed
second order partial derivative.
So now, I’ve got these three values here.
In order to evaluate at this point (-1,-2), I’m going to use a formula and I’m just
going to call it dxy.
So the formula I’m going to use for d of xy is going to be the second order partial
derivative with respect to x times the second order partial derivative with respect to y
minus the mixed second order partial derivative squared.
That’s the formula you’re going to use.
You’re always going to follow the same pattern here, second order partial derivative with
respect to x times second order partial derivative with respect to y minus the mixed second order
partial derivative squared.
So that’ll be my formula.
I’m going to evaluate and 2 times 2 gives me 4, minus negative 1 squared is a positive
1 so I end up with 3.
It doesn’t matter what we get here in terms of the specific value.
All that matters is whether or not this is greater than, less than or equal to zero.
In this case, obviously, we get 3 which is greater than zero so d of xy is greater than
zero.
When d of xy is greater than zero, we also need to look at the second order partial derivative
with respect to x.
So x squared.
We already know that that’s equal to 2 which is also greater than zero.
When these two are both greater than zero, that means that we have a local minimum at
the point at which we evaluated so local minimum at (-1,-2).
If d of xy was greater than zero and second derivative with respect to x was less than
zero, we’d have a local maximum.
If d of xy is less than zero, it doesn’t matter what the second order derivative with
respect to x is.
If d of xy is less than zero, then we’re looking at a sallow point.
And if d of xy is equal to zero, then the second derivative test is inconclusive and
we need to figure out another way to figure out the point (-1,-2).
But in this case, because they’re both greater than zero, we’ve got a local minimum at
(-1,-2).
So that’s it.
I hope this video helped you guys and I will see you in the next one.
Bye!