Question

Video Example EXAMPLE 3 Find the local maximum and minimum values and saddle points of f(x, y) = x4 + y4 − 4xy + 1. SOLUTION We first locate the critical points:

Answer 1

(0, 0) is a saddle point

(1, 1) is a local minimum

(-1, -1) is another point of local minimum

Explanation:

We first locate the critical points. In order to get the critical points we need to find the first derivatives and then set them to zero.

f(x, y) = x⁴ + y⁴ - 4 xy + 1

Find the first derivatives wrt to x and y

`f_(x)(x,y)` = 4x ³ - 4y --> (1)

`f_(y)(x,y)` = 4y³ - 4x --> (2)

Solve (1) for y

4x ³ - 4y = 0

x ³ = y

y = x ³ ---> (3)

Solve (2) for x

4y³ - 4x = 0

4y³ = 4x

y³ = x

x = y³ ----> (4)

Plug (3) into (2)

4y³ - 4x

4(x ³)³ - 4x = 0

4x⁹ - 4x = 0

4x (x ⁸ - 1) = 0

4x (x ⁴ - 1)(x⁴ + 1) = 0

4x (x ² - 1)(x ² + 1)(x ⁴ + 1) = 0

4x (x - 1)(x + 1)(x ² + 1)(x ⁴ + 1) = 0

So

x = 1 , -1 , 0

In order to find values of y, Plug each x value in (3)

For

x = 0

corresponding y value

y = x ³

y = 0 ³

y = 0

For

x = 1

y = x ³

y = 1 ³

y = 1

For

x = -1

y = x ³

y = (-1 )³

y = -1

Hence we get the critical points which are:

(0, 0)

(1, 1) and

(-1, -1)

Now for each critical point, we have to compute D(x,y)

For a critical point (x,y), D computed as:

D(x, y) =
`f_(xx)` (x, y) -
`f_(yy)` (x, y) - (
`f_(xy)` (x, y))²

After computing D(x,y) check:

If D(x, y) > 0 and
`f_(xx)` (x, y) > 0:

f(x, y) is a local minimum.

If D(x, y) > 0 and
`f_(xx)` (x, y) < 0:

f(x, y) is a local maximum.

If D(x, y) < 0:

then f(x, y) is a saddle point

Here first compute the second derivative in order to get
`f_(xx)` ,
`f_(yy)` and
`f_(xy)`

we have already computed:

`f_(x)(x,y)` = 4x ³ - 4y --> (1)

`f_(y)(x,y)` = 4y³ - 4x --> (2)

Now

`f_(xx)(x,y)` = 12x²

`f_(yy)(x,y)` = 12y²

`f_(xy)(x,y)` = -4

Compute D(x,y)

Critical Point (0, 0):

`D(0, 0) = f_(xx) (0,0) f_(yy)(0,0)-(f_(xy)(0,0))^(2)`

Putting values of
`f_(xx)(x,y)` ,
`f_(yy)(x,y)` and
`f_(xy)(x,y)` in above equation:

D(0,0) = 12(0)² * 12(0)² - (-4)²

= 0 * 0 -16

D(0,0) = -16

We know that if D < 0, the critical point f(x, y) is a saddle point.

D(0,0) < 0 because D(0,0) = -16

Hence (0, 0) is a saddle point

Compute D(x,y)

Critical Point (1, 1):

`D(1, 1) = f_(xx) (1,1) f_(yy)(1,1)-(f_(xy)(1,1))^(2)`

Putting values of
`f_(xx)(x,y)` ,
`f_(yy)(x,y)` and
`f_(xy)(x,y)` in above equation:

D(1,1) = 12(1)² * 12(1)² - (-4)²

= 12 * 12 - 16

D(1,1) = 128

We know that if D(x, y) > 0 and
`f_(xx)` (x, y) > 0 then f(x, y) is a local minimum.

D(1,1) > 0 because D(1,1) = 128

`f_(xx)` (x, y) > 0 because
`f_(xx)` (x, y) = 12

Hence (1,1) is the local minimum

Compute D(x,y)

Critical Point (-1, -1):

`D(-1, -1) = f_(xx) (-1,-1) f_(yy)(-1,-1)-(f_(xy)(-1,-1))^(2)`

Putting values of
`f_(xx)(x,y)` ,
`f_(yy)(x,y)` and
`f_(xy)(x,y)` in above equation:

D(-1,-1) = 12(-1)² * 12(-1)² - (-4)²

= 12 * 12 - 16

D(-1,-1) = 128

We know that if D(x, y) > 0 and
`f_(xx)` (x, y) > 0 then f(x, y) is a local minimum.

D(-1,-1) > 0 because D(-1,-1) = 128

`f_(xx)` (x, y) > 0 because
`f_(xx)` (x, y) = 12

Hence (-1,-1) is the local minimum