Answer :
Let's determine the locations of any relative extrema and identify any saddle points for the given function:
[tex]\[ f(x, y) = -3x^2 - 5xy - 3y^2 - 15x - 18y + 5 \][/tex]
### Step 1: Find the first partial derivatives
First, we need to calculate the partial derivatives of \( f \) with respect to \( x \) and \( y \):
[tex]\[ f_x = \frac{\partial f}{\partial x} \][/tex]
[tex]\[ f_x = -6x - 5y - 15 \][/tex]
[tex]\[ f_y = \frac{\partial f}{\partial y} \][/tex]
[tex]\[ f_y = -5x - 6y - 18 \][/tex]
### Step 2: Find critical points
To find the critical points, we set the first partial derivatives equal to zero and solve the resulting system of equations:
[tex]\[ -6x - 5y - 15 = 0 \][/tex]
[tex]\[ -5x - 6y - 18 = 0 \][/tex]
Solving this system of linear equations, we can use substitution or elimination method:
Multiply the first equation by 6 and the second by 5 to facilitate elimination of \( y \):
[tex]\[ -36x - 30y - 90 = 0 \quad \text{(Equation 1)} \][/tex]
[tex]\[ -25x - 30y - 90 = 0 \quad \text{(Equation 2)} \][/tex]
Now subtract Equation 2 from Equation 1:
[tex]\[ (-36x - 30y - 90) - (-25x - 30y - 90) = 0 \][/tex]
[tex]\[ -36x + 25x = 0 \][/tex]
[tex]\[ -11x = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Substitute \( x = 0 \) back into one of the original equations, say \( -6x - 5y - 15 = 0 \):
[tex]\[ -6(0) - 5y - 15 = 0 \][/tex]
[tex]\[ -5y - 15 = 0 \][/tex]
[tex]\[ -5y = 15 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the critical point is:
[tex]\[ (x, y) = (0, -3) \][/tex]
### Step 3: Find the second partial derivatives
Now, we need to find the second partial derivatives to determine the nature of the critical point:
[tex]\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} \][/tex]
[tex]\[ f_{xx} = -6 \][/tex]
[tex]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} \][/tex]
[tex]\[ f_{yy} = -6 \][/tex]
[tex]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} \][/tex]
[tex]\[ f_{xy} = -5 \][/tex]
### Step 4: Evaluate the Hessian determinant
The Hessian determinant \( D \) at the critical point \((0, -3)\) is given by:
[tex]\[ D = f_{xx}f_{yy} - (f_{xy})^2 \][/tex]
[tex]\[ D = (-6)(-6) - (-5)^2 \][/tex]
[tex]\[ D = 36 - 25 \][/tex]
[tex]\[ D = 11 \][/tex]
### Step 5: Classify the critical point
Since \( D > 0 \) and \( f_{xx} < 0 \):
- The critical point \((0, -3)\) is a relative maximum.
Therefore, the correct answer is:
B. There are no relative maxima.
[tex]\[ f(x, y) = -3x^2 - 5xy - 3y^2 - 15x - 18y + 5 \][/tex]
### Step 1: Find the first partial derivatives
First, we need to calculate the partial derivatives of \( f \) with respect to \( x \) and \( y \):
[tex]\[ f_x = \frac{\partial f}{\partial x} \][/tex]
[tex]\[ f_x = -6x - 5y - 15 \][/tex]
[tex]\[ f_y = \frac{\partial f}{\partial y} \][/tex]
[tex]\[ f_y = -5x - 6y - 18 \][/tex]
### Step 2: Find critical points
To find the critical points, we set the first partial derivatives equal to zero and solve the resulting system of equations:
[tex]\[ -6x - 5y - 15 = 0 \][/tex]
[tex]\[ -5x - 6y - 18 = 0 \][/tex]
Solving this system of linear equations, we can use substitution or elimination method:
Multiply the first equation by 6 and the second by 5 to facilitate elimination of \( y \):
[tex]\[ -36x - 30y - 90 = 0 \quad \text{(Equation 1)} \][/tex]
[tex]\[ -25x - 30y - 90 = 0 \quad \text{(Equation 2)} \][/tex]
Now subtract Equation 2 from Equation 1:
[tex]\[ (-36x - 30y - 90) - (-25x - 30y - 90) = 0 \][/tex]
[tex]\[ -36x + 25x = 0 \][/tex]
[tex]\[ -11x = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Substitute \( x = 0 \) back into one of the original equations, say \( -6x - 5y - 15 = 0 \):
[tex]\[ -6(0) - 5y - 15 = 0 \][/tex]
[tex]\[ -5y - 15 = 0 \][/tex]
[tex]\[ -5y = 15 \][/tex]
[tex]\[ y = -3 \][/tex]
So, the critical point is:
[tex]\[ (x, y) = (0, -3) \][/tex]
### Step 3: Find the second partial derivatives
Now, we need to find the second partial derivatives to determine the nature of the critical point:
[tex]\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} \][/tex]
[tex]\[ f_{xx} = -6 \][/tex]
[tex]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} \][/tex]
[tex]\[ f_{yy} = -6 \][/tex]
[tex]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} \][/tex]
[tex]\[ f_{xy} = -5 \][/tex]
### Step 4: Evaluate the Hessian determinant
The Hessian determinant \( D \) at the critical point \((0, -3)\) is given by:
[tex]\[ D = f_{xx}f_{yy} - (f_{xy})^2 \][/tex]
[tex]\[ D = (-6)(-6) - (-5)^2 \][/tex]
[tex]\[ D = 36 - 25 \][/tex]
[tex]\[ D = 11 \][/tex]
### Step 5: Classify the critical point
Since \( D > 0 \) and \( f_{xx} < 0 \):
- The critical point \((0, -3)\) is a relative maximum.
Therefore, the correct answer is:
B. There are no relative maxima.
