Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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.
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
