ong>Step-by-step explanation:

We can expand

f

to

f

(

x

,

y

)

=

x

y

−

x

2

y

−

x

y

2

. Next, find the PARTIAL derivatives and set them equal to zero.

∂

f

∂

x

=

y

−

2

x

y

−

y

2

=

y

(

1

−

2

x

−

y

)

=

0

∂

f

∂

y

=

x

−

x

2

−

2

x

y

=

x

(

1

−

x

−

2

y

)

=

0

Clearly,

(

x

,

y

)

=

(

0

,

0

)

,

(

1

,

0

)

,

and

(

0

,

1

)

are solutions to this system, and so are CRITICAL points of

f

. The other solution can be found from the system

1

−

2

x

−

y

=

0

,

1

−

x

−

2

y

=

0

. Solving the first equation for

y

in terms of

x

gives

y

=

1

−

2

x

, which can be plugged into the second equation to GET

1

−

x

−

2

(

1

−

2

x

)

=

0

⇒

−

1

+

3

x

=

0

⇒

x

=

1

3

. From this,

y

=

1

−

2

(

1

3

)

=

1

−

2

3

=

1

3

as well.

To test the nature of these critical points, we find second derivatives:

∂

2

f

∂

x

2

=

−

2

y

,

∂

2

f

∂

y

2

=

−

2

x

, and

∂

2

f

∂

x

∂

y

=

∂

2

f

∂

y

∂

x

=

1

−

2

x

−

2

y

.

The discriminant is therefore:

D

=

4

x

y

−

(

1

−

2

x

−

2

y

)

2

=

4

x

y

−

(

1

−

2

x

−

2

y

−

2

x

+

4

x

2

+

4

x

y

−

2

y

+

4

x

y

+

4

y

2

)

=

4

x

+

4

y

−

4

x

2

−

4

y

2

−

4

x

y

−

1

Plugging the first three critical points in gives:

D

(

0

,

0

)

=

−

1

<

0

,

D

(

1

,

0

)

=

4

−

4

−

1

=

−

1

<

0

, and

D

(

0

,

1

)

=

4

−

4

−

1

=

−

1

<

0

, making these points saddle points.

Plugging in the last critical point gives

D

(

1

3

,

1

3

)

=

4

3

+

4

3

−

4

9

−

4

9

−

4

9

−

1

=

1

3

>

0

. Also note that

∂

2

f

∂

x

2

(

1

3

,

1

3

)

=

−

2

3

<

0

. Therefore,

(

1

3

,

1

3

)

is a location of a LOCAL maximum value of

f

. You can check that the local maximum value itself is

f

(

1

3

,

1

3

)

=

1

27

.

Below is a picture of the contour map (of level curves) of

f

(the curves where the OUTPUT of

f

is constant), along with the 4 critical points of

f

.