Question

The function $f(x,y) = x^2 + y^2 - xy - x - y + 5$ has

• A. Maximum at (1,1)
• B. Saddle point at (1,1)
• C. Minimum at (1,1)
• D. Minimum at (1,2)

Hint:

Use the second derivative test for functions of two variables: \( D = f_{xx} f_{yy} - (f_{xy})^2 \); if \( D > 0 \) and \( f_{xx} > 0 \), it’s a minimum.

Answer 1

The Correct Option is C

Step 1: Find the critical points. 
To find the critical points of \( f(x, y) = x^2 + y^2 - xy - x - y + 5 \), compute the partial derivatives and set them to zero: \[ f_x = \frac{\partial f}{\partial x} = 2x - y - 1, \] \[ f_y = \frac{\partial f}{\partial y} = 2y - x - 1. \] Set \( f_x = 0 \) and \( f_y = 0 \): \[ 2x - y - 1 = 0 \quad \text{(1)}, \] \[ 2y - x - 1 = 0 \quad \text{(2)}. \] Solve equation (1) for \( y \): \[ y = 2x - 1. \] Substitute into equation (2): \[ 2(2x - 1) - x - 1 = 0 \implies 4x - 2 - x - 1 = 0 \implies 3x - 3 = 0 \implies x = 1. \] Then: \[ y = 2(1) - 1 = 1. \] The critical point is \( (x, y) = (1, 1) \). 
Step 2: Classify the critical point using the second derivative test. 
Compute the second partial derivatives: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2, \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2, \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -1. \] The discriminant for the second derivative test is: \[ D = f_{xx} f_{yy} - (f_{xy})^2. \] At \( (1, 1) \): \[ D = (2)(2) - (-1)^2 = 4 - 1 = 3. \] Since \( D>0 \), check \( f_{xx} \):
\( f_{xx} = 2>0 \), indicating a local minimum.
Thus, \( (1, 1) \) is a local minimum. 
Step 3: Check other points in the options. 
Option (4) suggests a minimum at \( (1, 2) \). Evaluate the partial derivatives at \( (1, 2) \): \[ f_x(1, 2) = 2(1) - 2 - 1 = -1 \neq 0, \] \[ f_y(1, 2) = 2(2) - 1 - 1 = 2 \neq 0. \] Since the partial derivatives are not zero, \( (1, 2) \) is not a critical point, so it cannot be a minimum. 
Step 4: Evaluate the options. 
(1) Maximum at (1,1): Incorrect, as \( f_{xx}>0 \) and \( D>0 \), indicating a minimum, not a maximum. Incorrect.
(2) Saddle point at (1,1): Incorrect, as \( D>0 \), indicating no saddle point (\( D<0 \) for a saddle point). Incorrect.
(3) Minimum at (1,1): Correct, as the second derivative test confirms a local minimum at \( (1, 1) \). Correct.
(4) Minimum at (1,2): Incorrect, as \( (1, 2) \) is not a critical point. Incorrect. 
Step 5: Select the correct answer. 
The function has a minimum at \( (1, 1) \), matching option (3).