For us to calculate all of the critical points
of the function,
then, determine if each point represents
a relative maximum, relative minimum,
or a saddle point.
For the first type,
we will determine the critical points or critical numbers,
by determining where the first order
of partial derivatives are equal to 0 or undefined,
and then step two, perform the second partials test.
This begin by determining all the partial derivatives
that we need for step one and step two.
Let's first determine the partial of f with respect to x,
which means we differentiate f of x, comma, y
with respect to x treating y as a constant
because we're differentiating with respect to x
treating y as a constant.
It might be helpful to write this last term as,
6 divided by y times x to the power of -1.
And now, differentiating with respect to x
treating y as a constant,
we have the partial of f
with respect to x is equal to 6x to the 5th
plus 0
and then, to differentiate 6 divided by xy
with respect to x,
we multiply by -1
and then, subtract 1 from the exponent of x,
which it gives us -6 divided by y
times x to the power of -2
or -6 divided by x squared, y.
Remember if we have x to the power of -2,
we move it down to the denominator,
it changes the sign of the exponent.
And now, let's find the partial of f with respect to y.
So, now we differentiate f of x, comma, y,
with respect to y, treating x as a constant.
And therefore, it might be helpful to rewrite the third term
as +6 divided by x,
y to the power of -1.
Differentiating with respect to y,
the derivative of f to 6 is 0,
+6y to the 5th,
giving us 6y to the 5th,
And then, differentiating 6 divided by x,
y to the power of -1 with respect to y,
we multiply by -1
and then, subtract 1 on the exponent of y
which gives us -6 divided by x,
y to the power of -2.
Or simplifying -6 divided by xy squared.
Let's also find the second order of partials that we need
to perform the second partials test.
Let's find the second order partials with respect to x,
which means we differentiate
the first order of partials with respect to x
with respect to x again.
So again, it might be helpful write the second term
as -6 divided by y,
times x to the power of -2.
To differentiate the respect to x,
we have 30x to the 4th
and then, for the second term,
we multiply by -2,
which gives us +2 times 6 divided by a y,
x to the power of -3,
2 times 6 is 12.
So, we have +12 divided by x cubed, y
And, now let's find the second order mixed partial,
the partial of f with respect to x,
then, with respect to y.
To find this partial derivative,
it'll be helpful to write the second term
of the first order of partial with respect to x
as -6 divided by x squared,
y to the power of -1.
Differentiating with respect to y,
we multiply by -1
which becomes +6 divided by x squared
times y to the power of -2,
giving us 6 divided by x squared, y squared.
And then, we also have to find the second order partial
with respect to y,
which means we differentiate the first order of partial
with respect to y, with respect to y again.
So, looking at the second term,
we can think of this as -6, divided by x,
y to the power of -2.
Differentiating with respect to y,
we have 30y to the 4th,
and then, for the second term,
we multiply by -2
which gives us +2 times 6,
which it'd be 12 divided by x
times y to the power of -3,
giving us +12 divided by xy cubed.
Now, that we have all the partial derivatives
that we need,
the next step is find their critical numbers,
which are the x and y values,
where the first order of partials are both equal to 0.
Let's work on this on the next slide.
So, we need to solve this system of equations,
6x to the 5th,
6 divided by x squared y equals 0
and 6y to the 5th,
6 divided by xy squared equals 0.
Let's begin by clearing the fractions from both equations.
For the first equation,
we multiply both sides of the equation by x squared y
which gives us 6x to the 7th, y,
6 equals 0.
Let's go ahead and solve this equation
for x to the 7th, y.
So, let's go ahead and add 6 to both sides,
and then, divide both sides by 6,
which gives us x to the 7th, y is equal to 1.
And now, let's go to the second equation
and multiply both sides of the equation by xy squared.
Simplifying, we have 6xy to the 7th,
6 equals 0.
Let's go ahead and solve this equation
for xy to the 7th.
So, let's add 6 to both sides,
divide both sides by 6
which gives us xy to the 7th equals 1.
So, we are still trying to solve the system of equations
where now the two equations are
x to the 7th, y equals 1
and xy to the 7th equals 1
but we can probably stop here
and solve this by inspection
where both equations are true
when x equals 1 and y equals 1,
so, one critical point would be 1, comma, 1, comma,
the z-coordinate would be f of 1, comma, 1,
and both equations are also true
when both x and y are -1.
And, therefore, second critical point is
1, comma, -1, comma, f of -1, comma, -1.
Let's go ahead and find the function values
of z-coordinates on the next slide.
I've already set this step to save some time.
Notice how both f of 1, comma, 1
and f of -1, comma, -1
are both equal to 8.
Remember we do use the original function
to find these z-coordinates.
So, the 2 critical points are
1, comma, 1, comma, 8.
as well as -1, comma, -1, comma, 8.
The next step is to perform the second partials test
which we will do on the next slide.
Now, I've already set a lot of this up to save some time.
I've copied the second order of partials that we need
in the upper right-hand corner.
To find d at the point,
1, comma, 1, comma, 8,
we evaluate the second order of partials
from the formula for d
where x and y are both +1.
In simplifying, we get d equals 1728.
Next, we find d for the point,
1, comma, -1, comma, 8,
so we determine d when x is equal to -1,
and y is equal to -1
which also gives us 1728.
So, as I said both critical points,
d is greater than 0 or positive,
the next step is find the sign of the second order
of partial with respect to x
at 1, comma, 1,
and -1, comma, -1.
And again, notice both are also positive
or greater than 0.
So, looking at our notes because d is greater than 0,
and the second order of partial with respect to x
is also greater than 0,
we have relative minimums at both critical points.
And, let's verify this graphically.
So, here we see the graph of the surface,
notice how we have one low point here,
if we keep rotating the surface,
we have another low point on the other side
which we see here.
So, this graph does verify
that we have two relative minimums.
Let's go ahead and summarize our findings.
Again, we have a relative minimum
at 1, comma, 1, comma, 8,
as well as a -1, comma, -1, comma, 8.
Another way to express this information would be to say,
the given function has a relative minimum of the z-value
or function value of 8 at the locations
of 1, comma, 1
and -1, comma, -1.
I hope you found this helpful.