Using the Second Partials Test
Find the relative extrema of f(x,y)=-x^(3)+4xy-2y^(2)+1
fx = 3 x^2 + 4 y
level if x = +/-(4y/3)^.5
fxx = 6 x
fxy = 4
fy = 4 x - 4 y level when x = y
fyy = -4
fyx = 4
extreme only when x = y = 0
H matrix
6 x 4
4 -4
det = -6x - 4
if x = 0, det = -4
if det <0 then (0,0) is a "saddle point"
To find the relative extrema of the function f(x, y) = -x^3 + 4xy - 2y^2 + 1 using the Second Partials Test, follow these steps:
Step 1: Calculate the first-order partial derivatives of f(x, y) with respect to x and y.
∂f/∂x = -3x^2 + 4y
∂f/∂y = 4x - 4y
Step 2: Calculate the second-order partial derivatives of f(x, y) with respect to x and y.
∂^2f/∂x^2 = -6x
∂^2f/∂y^2 = -4
∂^2f/∂x∂y = 4
∂^2f/∂y∂x = 4 (since the order of differentiation does not matter for continuous partial derivatives)
Step 3: Evaluate the second-order derivatives at each critical point.
To find the critical points, set the first-order partial derivatives equal to zero and solve for x and y:
∂f/∂x = -3x^2 + 4y = 0
∂f/∂y = 4x - 4y = 0
Solving these equations simultaneously, we get:
-3x^2 + 4y = 0 ...(Equation 1)
4x - 4y = 0 ...(Equation 2)
Simplifying Equation 2, we get:
x - y = 0 ...(Equation 3)
Substituting Equation 3 into Equation 1:
-3x^2 + 4(x) = 0
-3x^2 + 4x = 0
x(-3x + 4) = 0
So, either x = 0 or -3x +4 = 0.
If x = 0, then from Equation 3, y = x = 0.
If -3x + 4 = 0, then x = 4/3. Substituting into Equation 3, y = x = 4/3.
Therefore, we have two critical points: (0, 0) and (4/3, 4/3).
Step 4: Determine the nature of each critical point by evaluating the second-order partial derivatives.
For the critical point (0, 0):
Substitute x = 0 and y = 0 into the second-order partial derivatives:
∂^2f/∂x^2 = -6(0) = 0
∂^2f/∂y^2 = -4
∂^2f/∂x∂y = 4
∂^2f/∂y∂x = 4
Calculate the discriminant D = (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2:
D = (0)(-4) - (4)(4) = -16
Since D < 0 and ∂^2f/∂x^2 = 0, we can conclude that there is no determination of the nature of the critical point (0, 0) using the Second Partials Test.
For the critical point (4/3, 4/3):
Substitute x = 4/3 and y = 4/3 into the second-order partial derivatives:
∂^2f/∂x^2 = -6(4/3) = -8
∂^2f/∂y^2 = -4
∂^2f/∂x∂y = 4
∂^2f/∂y∂x = 4
Calculate the discriminant D = (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)^2:
D = (-8)(-4) - (4)(4) = 16 - 16 = 0
Since D = 0, the Second Partials Test is inconclusive for determining the nature of the critical point (4/3, 4/3).
Step 5: Interpret the results.
From our analysis using the Second Partials Test, both critical points (0, 0) and (4/3, 4/3) are inconclusive, and we cannot determine their nature (whether they are relative maxima, minima, or saddle points) based on the second-order partial derivatives.
Therefore, we need to further analyze the function and consider other methods to determine the nature of the critical points (0, 0) and (4/3, 4/3).
To find the relative extrema of the function f(x, y) = -x^3 + 4xy - 2y^2 + 1 using the Second Partials Test, follow these steps:
Step 1: Find the first-order partial derivatives of f(x, y) with respect to x and y.
- To find ∂f/∂x, differentiate f(x, y) with respect to x while treating y as a constant.
- To find ∂f/∂y, differentiate f(x, y) with respect to y while treating x as a constant.
Differentiating f(x, y) = -x^3 + 4xy - 2y^2 + 1 with respect to x:
∂f/∂x = -3x^2 + 4y
Differentiating f(x, y) = -x^3 + 4xy - 2y^2 + 1 with respect to y:
∂f/∂y = 4x - 4y
Step 2: Find the second-order partial derivatives of f(x, y) with respect to x and y.
- To find ∂^2f/∂x^2, differentiate ∂f/∂x with respect to x.
- To find ∂^2f/∂y^2, differentiate ∂f/∂y with respect to y.
- To find ∂^2f/∂x∂y, differentiate ∂f/∂x with respect to y.
- To find ∂^2f/∂y∂x, differentiate ∂f/∂y with respect to x.
Differentiating ∂f/∂x = -3x^2 + 4y with respect to x:
∂^2f/∂x^2 = -6x
Differentiating ∂f/∂y = 4x - 4y with respect to y:
∂^2f/∂y^2 = -4
Differentiating ∂f/∂x = -3x^2 + 4y with respect to y:
∂^2f/∂x∂y = 4
Differentiating ∂f/∂y = 4x - 4y with respect to x:
∂^2f/∂y∂x = 4
Step 3: Evaluate the second-order partial derivatives at the critical points.
- Find the critical points by setting the first-order partial derivatives equal to zero and solving for x and y.
- Plug the critical point values into the second-order partial derivatives.
Setting ∂f/∂x = -3x^2 + 4y equal to zero:
-3x^2 + 4y = 0
Setting ∂f/∂y = 4x - 4y equal to zero:
4x - 4y = 0
Solving these equations, we have:
x = 0 and y = 0
Evaluating the second-order partial derivatives at (x = 0, y = 0):
∂^2f/∂x^2 = -6(0) = 0
∂^2f/∂y^2 = -4
∂^2f/∂x∂y = 4
∂^2f/∂y∂x = 4
Step 4: Apply the Second Partials Test.
- The Second Partials Test states that if the second-order partial derivatives satisfy the following conditions:
a) The determinant D = (∂^2f/∂x^2)(∂^2f/∂y^2) - (∂^2f/∂x∂y)(∂^2f/∂y∂x) is positive,
b) (∂^2f/∂x^2) is negative for the critical point,
then the critical point is a relative maximum.
Evaluating the conditions for (x = 0, y = 0):
D = (0)(-4) - (4)(4) = -16
(∂^2f/∂x^2) = 0
Since the determinant D is negative and (∂^2f/∂x^2) is zero, the Second Partials Test is inconclusive.
Therefore, using the Second Partials Test, we cannot determine the relative extrema of f(x, y) = -x^3 + 4xy - 2y^2 + 1 at the critical point (0, 0). Other methods, such as the First Derivative Test or studying the behavior of the function around the critical point, may be used to determine the extrema.