Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Question

Make all obvious simplifications.

### Problem Statement **Question 8** **A.** Suppose that \( z = f(x, y) \) is a function for which: - \( f(3, -1) = 16 \), - \( f_x(3, -1) = 0 \), - \( f_y(3, -1) = 0 \), - \( f_{xx}(3, -1) = -3 \), - \( f_{yy}(3, -1) = -7 \), and - \( f_{xy}(3, -1) = f_{yx}(3, -1) = 4 \). Does this data say that \( f(3, -1) = 16 \) is a maximum, minimum, or neither? **Briefly** explain your reasoning. **B.** Suppose I had changed the mixed partial to \( f_{xy}(3, -1) = f_{yx}(3, -1) = 5 \). What would your answer be in this case? ### Explanation This question involves using the second derivative test for functions of two variables to determine the local extremum at a given point. #### Second Derivative Test 1. Calculate the discriminant: \[ D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - [f_{xy}(x_0, y_0)]^2 \] 2. Evaluate at the point \( (3, -1) \): - For Part A: \[ D = (-3) \cdot (-7) - 4^2 = 21 - 16 = 5 \] - Since \( D > 0 \) and \( f_{xx}(3, -1) < 0 \), \( f(3, -1) = 16 \) is a local maximum. 3. For Part B, substitute the changed mixed partials: \[ D = (-3) \cdot (-7) - 5^2 = 21 - 25 = -4 \] - Since \( D < 0 \), the point \( (3, -1) \) is a saddle point, meaning neither a local maximum nor a local minimum.
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Transcribed Image Text:### Problem Statement **Question 8** **A.** Suppose that \( z = f(x, y) \) is a function for which: - \( f(3, -1) = 16 \), - \( f_x(3, -1) = 0 \), - \( f_y(3, -1) = 0 \), - \( f_{xx}(3, -1) = -3 \), - \( f_{yy}(3, -1) = -7 \), and - \( f_{xy}(3, -1) = f_{yx}(3, -1) = 4 \). Does this data say that \( f(3, -1) = 16 \) is a maximum, minimum, or neither? **Briefly** explain your reasoning. **B.** Suppose I had changed the mixed partial to \( f_{xy}(3, -1) = f_{yx}(3, -1) = 5 \). What would your answer be in this case? ### Explanation This question involves using the second derivative test for functions of two variables to determine the local extremum at a given point. #### Second Derivative Test 1. Calculate the discriminant: \[ D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - [f_{xy}(x_0, y_0)]^2 \] 2. Evaluate at the point \( (3, -1) \): - For Part A: \[ D = (-3) \cdot (-7) - 4^2 = 21 - 16 = 5 \] - Since \( D > 0 \) and \( f_{xx}(3, -1) < 0 \), \( f(3, -1) = 16 \) is a local maximum. 3. For Part B, substitute the changed mixed partials: \[ D = (-3) \cdot (-7) - 5^2 = 21 - 25 = -4 \] - Since \( D < 0 \), the point \( (3, -1) \) is a saddle point, meaning neither a local maximum nor a local minimum.
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