We're asked to use the method
of Lagrange multipliers to find the points
that represent the relative minimum and relative maximum
for F of X, Y equals XY,
subject to the constraint Y equals nine minus X squared.
We will assume the conditions of Lagrange's Theorem
have been met and therefore if F of X, Y has a max or min,
subject to G of X, Y equals zero,
then the gradient of F is equal to lambda times the gradient
of G and G of X, Y equals zero,
which results in the system of equations below
that we need to solve for X, Y and lambda.
To begin, we are given F of X, Y is equal to XY.
To write the constraint G of X, Y equals zero,
we need to set one side of the equation
of Y equals nine minus X squared to zero.
Let's go ahead and add X squared to both sides
and also subtract nine, which gives us G of X, Y is equal
to Y plus X squared minus nine.
And now let's work on setting up the first equation given
by the partial of F with respect to X equals lambda times
the partial of G with respect to X.
To find the partial of F with respect to X,
we differentiate XY with respect to X, treating Y
as a constant, which gives us one times Y or Y.
We have Y equals lambda times a partial
of G with respect to X.
To find the partial of G with respect to X,
we differentiate Y plus X squared minus nine with respect
to X, treating Y as a constant, which gives us zero
plus 2X minus zero or just 2X.
The next equation is the partial of F with respect to Y,
equals lambda times the partial of G with respect to Y.
To find the partial of F with respect to Y,
we differentiate XY with respect to Y, treating X
as a constant, which gives us X times one or X,
equals lambda times, to find the partial of G with respect
to Y, we differentiate Y plus X squared minus nine
with respect to Y, treating X as a constant
which gives us one plus zero minus zero or just one.
The third equation of the system is G of X, Y equals zero,
which is Y plus X squared minus nine equals zero.
To begin solving the system, let's go ahead
and solve these first two equations for lambda.
For the first equation, we would divide both sides by 2X
where lambda is equal to Y divided by 2X.
And for the second equation, we have lambda equals X.
Note, both equations are solved for lambda
which means Y divided by 2X must equal X.
Let's go and take a look at that equation.
Again, if lambda equals Y divided by 2X,
and lambda's also equal to X,
we can say Y divided by 2X is equal to X.
And now let's go ahead and solve this equation
for Y by multiplying both sides by 2X.
Simplifying, we have Y equals 2X squared.
And notice we have a Y in our third equation.
We have Y plus X squared minus nine equals zero.
So now, if we substitute 2X squared for Y,
we will have one equation
with one variable that we can solve for X.
So performing the substitution,
we would have 2X squared plus X squared.
Let's go ahead and add nine to both sides,
giving us equals nine.
Combining like terms, we have 3X squared equals nine,
dividing both sides by three,
we have X squared is equal to three,
and now we take the square root of both sides
of the equation and we are going to have two solution,
a positive and negative solution.
We have X equals plus or minus the square of three.
Now that we know the values of X, we can determine lambda,
since lambda is equal to X, as well as Y.
And we are going to have two points.
Let's go ahead and start running these out.
One point is going to have an X coordinate
of positive square root of three
and the second point is going to have an X coordinate
of negative square root of three.
And again, because lambda is equal to X,
we know lambda is also equal
to plus or minus square root of three.
So when X is positive, lambda is positive.
When X is negative, lambda is negative.
And now to determine Y using this first equation here.
Y is equal to lambda times 2X.
So we have Y equals, let's use positive square
of three first, we have square root
of three times two square root of three.
Well, square root of three times square of three is three.
This part is equal to six.
So when X is positive square root of three, Y is six.
And now let's find Y when X is negative square root
of three and so is lambda.
We have Y equals negative square root of three,
times two, times negative square root of three
or negative two square root of three
but is also equal to positive six.
So the second point when X is negative square root of three,
Y is also positive six.
And now we need to find the Z coordinates
or the function values when X is square root of three
and Y is six as well
as when X is negative square root of three and Y is six.
And since the function is F of X, Y equals X times Y,
the Z coordinate for this point is just square root
of three times six or six square root of three.
And for this point, we have negative six square root
of three.
So obviously the larger function value is six square root
of three and the least function value is negative six square
root of three.
So the first point represents the relative maximum
where the location is often given by the ordered pair X, Y
and the function value
or Z coordinate in this case is the relative max.
And for the second point, the location is given
by negative square root of three, six.
And the relative minimum function value is
negative six square root of three.
And let's go ahead and verify this graphically.
F of X, Y is the graph of the surface
and then the space curve is the intersection
of the constraint and the surface.
And notice how the two points that we found,
one is a relative high point or a local maximum here
and the point down here is a relative
or local minimum verifying our work is correct.
We did find the relative extrema.
Also notice how the space curve does go up indefinitely
as well as down indefinitely
which is why we have relative extrema here,
not absolute extrema.
I hope you found this helpful.