Answer :
To find the values of \( x \) and \( y \) that satisfy \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) for the function \( f(x, y) = 7x^2 + 9y^2 + 2xy + 32x - 6 \), we will follow a detailed, step-by-step method:
### Step 1: Calculate the partial derivatives
First, we compute the partial derivatives of \( f \) with respect to \( x \) and \( y \).
For the partial derivative with respect to \( x \):
[tex]\[ f_x(x, y) = \frac{\partial}{\partial x} (7x^2 + 9y^2 + 2xy + 32x - 6) \][/tex]
Applying the rules of differentiation:
[tex]\[ f_x(x, y) = 14x + 2y + 32 \][/tex]
For the partial derivative with respect to \( y \):
[tex]\[ f_y(x, y) = \frac{\partial}{\partial y} (7x^2 + 9y^2 + 2xy + 32x - 6) \][/tex]
Applying the rules of differentiation:
[tex]\[ f_y(x, y) = 2x + 18y \][/tex]
### Step 2: Set the partial derivatives equal to zero
We need to find the values of \( x \) and \( y \) such that both partial derivatives are zero:
[tex]\[ f_x(x, y) = 14x + 2y + 32 = 0 \][/tex]
[tex]\[ f_y(x, y) = 2x + 18y = 0 \][/tex]
### Step 3: Solve the system of equations
We solve the system of linear equations:
1. \( 14x + 2y + 32 = 0 \)
2. \( 2x + 18y = 0 \)
First, we simplify the second equation:
[tex]\[ 2x + 18y = 0 \implies x = -9y \][/tex]
Next, we substitute \( x = -9y \) into the first equation:
[tex]\[ 14(-9y) + 2y + 32 = 0 \][/tex]
[tex]\[ -126y + 2y + 32 = 0 \][/tex]
[tex]\[ -124y + 32 = 0 \][/tex]
[tex]\[ -124y = -32 \][/tex]
[tex]\[ y = \frac{32}{124} = \frac{8}{31} \][/tex]
With \( y = \frac{8}{31} \), we substitute back to find \( x \):
[tex]\[ x = -9y = -9 \left(\frac{8}{31}\right) = -\frac{72}{31} \][/tex]
### Step 4: Present the solution
Therefore, the values of \( x \) and \( y \) that satisfy both \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) are:
[tex]\[ x = -\frac{72}{31} \quad \text{and} \quad y = \frac{8}{31} \][/tex]
The critical point of the function [tex]\( f(x, y) \)[/tex] is [tex]\(\left(-\frac{72}{31}, \frac{8}{31}\right)\)[/tex].
### Step 1: Calculate the partial derivatives
First, we compute the partial derivatives of \( f \) with respect to \( x \) and \( y \).
For the partial derivative with respect to \( x \):
[tex]\[ f_x(x, y) = \frac{\partial}{\partial x} (7x^2 + 9y^2 + 2xy + 32x - 6) \][/tex]
Applying the rules of differentiation:
[tex]\[ f_x(x, y) = 14x + 2y + 32 \][/tex]
For the partial derivative with respect to \( y \):
[tex]\[ f_y(x, y) = \frac{\partial}{\partial y} (7x^2 + 9y^2 + 2xy + 32x - 6) \][/tex]
Applying the rules of differentiation:
[tex]\[ f_y(x, y) = 2x + 18y \][/tex]
### Step 2: Set the partial derivatives equal to zero
We need to find the values of \( x \) and \( y \) such that both partial derivatives are zero:
[tex]\[ f_x(x, y) = 14x + 2y + 32 = 0 \][/tex]
[tex]\[ f_y(x, y) = 2x + 18y = 0 \][/tex]
### Step 3: Solve the system of equations
We solve the system of linear equations:
1. \( 14x + 2y + 32 = 0 \)
2. \( 2x + 18y = 0 \)
First, we simplify the second equation:
[tex]\[ 2x + 18y = 0 \implies x = -9y \][/tex]
Next, we substitute \( x = -9y \) into the first equation:
[tex]\[ 14(-9y) + 2y + 32 = 0 \][/tex]
[tex]\[ -126y + 2y + 32 = 0 \][/tex]
[tex]\[ -124y + 32 = 0 \][/tex]
[tex]\[ -124y = -32 \][/tex]
[tex]\[ y = \frac{32}{124} = \frac{8}{31} \][/tex]
With \( y = \frac{8}{31} \), we substitute back to find \( x \):
[tex]\[ x = -9y = -9 \left(\frac{8}{31}\right) = -\frac{72}{31} \][/tex]
### Step 4: Present the solution
Therefore, the values of \( x \) and \( y \) that satisfy both \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \) are:
[tex]\[ x = -\frac{72}{31} \quad \text{and} \quad y = \frac{8}{31} \][/tex]
The critical point of the function [tex]\( f(x, y) \)[/tex] is [tex]\(\left(-\frac{72}{31}, \frac{8}{31}\right)\)[/tex].
