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For the function f(x,y) = 2x4 - 3x2y + y2 has
- a)maximum at (0, 0)
- b)minimum at (0, 0)
- c)neither maxima nor minima at (0, 0)
- d)doubtful case at (0, 0) always
Correct answer is option 'C'. Can you explain this answer?
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For the function f(x,y) = 2x4 - 3x2y + y2hasa)maximum at (0, 0)b)minim...
Explanation:
To determine whether the given function has a maximum, minimum, or neither at the point (0, 0), we need to find the critical points and examine the behavior of the function near those points.
Step 1: Find the Partial Derivatives
To find the critical points, we need to take the partial derivatives of the function with respect to x and y.
The partial derivative with respect to x is:
∂f/∂x = 8x^3 - 6xy
The partial derivative with respect to y is:
∂f/∂y = -3x^2
Step 2: Set the Partial Derivatives to Zero
To find the critical points, we need to set the partial derivatives equal to zero and solve for x and y.
Setting ∂f/∂x = 0:
8x^3 - 6xy = 0
Factor out x:
x(8x^2 - 6y) = 0
Setting ∂f/∂y = 0:
-3x^2 = 0
This equation has only one solution, x = 0.
Step 3: Analyze the Critical Points
We have found the critical points at (0, 0).
To determine whether these critical points are maximum, minimum, or neither, we can use the second partial derivative test.
The second partial derivatives are:
∂^2f/∂x^2 = 24x^2 - 6y
∂^2f/∂y^2 = 0
Step 4: Apply the Second Partial Derivative Test
Evaluate the second partial derivatives at the critical point (0, 0).
∂^2f/∂x^2 = 24(0)^2 - 6(0) = 0
∂^2f/∂y^2 = 0
Since the second partial derivatives are both zero, the second partial derivative test is inconclusive. Therefore, we cannot determine whether the critical point (0, 0) is a maximum, minimum, or neither.
Conclusion:
Based on the analysis, the function f(x, y) = 2x^4 - 3x^2y does not have a maximum or minimum at the point (0, 0). The critical point (0, 0) is neither a maximum nor a minimum.
To determine whether the given function has a maximum, minimum, or neither at the point (0, 0), we need to find the critical points and examine the behavior of the function near those points.
Step 1: Find the Partial Derivatives
To find the critical points, we need to take the partial derivatives of the function with respect to x and y.
The partial derivative with respect to x is:
∂f/∂x = 8x^3 - 6xy
The partial derivative with respect to y is:
∂f/∂y = -3x^2
Step 2: Set the Partial Derivatives to Zero
To find the critical points, we need to set the partial derivatives equal to zero and solve for x and y.
Setting ∂f/∂x = 0:
8x^3 - 6xy = 0
Factor out x:
x(8x^2 - 6y) = 0
Setting ∂f/∂y = 0:
-3x^2 = 0
This equation has only one solution, x = 0.
Step 3: Analyze the Critical Points
We have found the critical points at (0, 0).
To determine whether these critical points are maximum, minimum, or neither, we can use the second partial derivative test.
The second partial derivatives are:
∂^2f/∂x^2 = 24x^2 - 6y
∂^2f/∂y^2 = 0
Step 4: Apply the Second Partial Derivative Test
Evaluate the second partial derivatives at the critical point (0, 0).
∂^2f/∂x^2 = 24(0)^2 - 6(0) = 0
∂^2f/∂y^2 = 0
Since the second partial derivatives are both zero, the second partial derivative test is inconclusive. Therefore, we cannot determine whether the critical point (0, 0) is a maximum, minimum, or neither.
Conclusion:
Based on the analysis, the function f(x, y) = 2x^4 - 3x^2y does not have a maximum or minimum at the point (0, 0). The critical point (0, 0) is neither a maximum nor a minimum.
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For the function f(x,y) = 2x4 - 3x2y + y2hasa)maximum at (0, 0)b)minimum at (0, 0)c)neither maxima nor minima at (0, 0)d)doubtful case at (0, 0) alwaysCorrect answer is option 'C'. Can you explain this answer?
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