Question:
Consider the function
\[
f(x, y) = x^3 - y^3 - 3x^2 + 3y^2 + 7, \, x, y \in \mathbb{R}.
\]
Then the local minimum (\( m \)) and the local maximum (\( M \)) of \( f \) are given by
Consider the function
\[
f(x, y) = x^3 - y^3 - 3x^2 + 3y^2 + 7, \, x, y \in \mathbb{R}.
\]
Then the local minimum (\( m \)) and the local maximum (\( M \)) of \( f \) are given by
Show Hint
For multivariable functions, use partial derivatives and the second derivative test to determine the nature of critical points.
Updated On: Nov 18, 2025
- \( m = 3, M = 7 \)
- \( m = 4, M = 11 \)
- \( m = 7, M = 11 \)
- \( m = 3, M = 11 \)
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The Correct Option is D
Solution and Explanation
Step 1: Find the critical points.
To find the critical points of \( f(x, y) \), we take the partial derivatives with respect to \( x \) and \( y \), and set them equal to 0: \[ f_x = 3x^2 - 6x, \quad f_y = -3y^2 + 6y \] Solve these equations to find the critical points.
Step 2: Check the nature of the critical points.
We use the second derivative test to classify the critical points as local minima or maxima. After calculation, we find that \( m = 3 \) and \( M = 11 \).
To find the critical points of \( f(x, y) \), we take the partial derivatives with respect to \( x \) and \( y \), and set them equal to 0: \[ f_x = 3x^2 - 6x, \quad f_y = -3y^2 + 6y \] Solve these equations to find the critical points.
Step 2: Check the nature of the critical points.
We use the second derivative test to classify the critical points as local minima or maxima. After calculation, we find that \( m = 3 \) and \( M = 11 \).
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