(0, 0) is a saddle point
(1, 1) is a local minimum
(-1, -1) is another point of local minimum
Step-by-step 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
= 4x
³ - 4y --> (1)
= 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) = 
(x, y) - 
(x, y) - (
(x, y))²
After computing D(x,y) check:
If D(x, y) > 0 and (x, y) > 0:
f(x, y) is a local minimum.
If D(x, y) > 0 and (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 , and
we have already computed:
= 4x
³ - 4y --> (1)
= 4y³ - 4x --> (2)
Now
= 12x²
= 12y²
= -4
Compute D(x,y)
Critical Point (0, 0):
Putting values of , and 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):
Putting values of , and 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 (x, y) > 0 then f(x, y) is a local minimum.
D(1,1) > 0 because D(1,1) = 128
(x, y) > 0 because (x, y) = 12
Hence (1,1) is the local minimum
Compute D(x,y)
Critical Point (-1, -1):
Putting values of , and 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 (x, y) > 0 then f(x, y) is a local minimum.
D(-1,-1) > 0 because D(-1,-1) = 128
(x, y) > 0 because (x, y) = 12
Hence (-1,-1) is the local minimum
