Answer :
To find the critical points of the function \( f(x, y) = 7 x^2 + 9 y^2 + 2 x y + 32 x - 6 \), we need to solve for \( x \) and \( y \) such that both partial derivatives \( f_x(x, y) = 0 \) and \( f_y(x, y) = 0 \).
First, we find the partial derivatives of the function \( f(x, y) \):
1. Compute the partial derivative with respect to \( x \):
[tex]\[ f_x(x, y) = \frac{\partial}{\partial x} (7 x^2 + 9 y^2 + 2 x y + 32 x - 6) \][/tex]
[tex]\[ f_x(x, y) = 14 x + 2 y + 32 \][/tex]
2. Compute the partial derivative with respect to \( y \):
[tex]\[ f_y(x, y) = \frac{\partial}{\partial y} (7 x^2 + 9 y^2 + 2 x y + 32 x - 6) \][/tex]
[tex]\[ f_y(x, y) = 18 y + 2 x \][/tex]
Next, we set the partial derivatives to zero to find the critical points:
[tex]\[ 14 x + 2 y + 32 = 0 \][/tex]
[tex]\[ 18 y + 2 x = 0 \][/tex]
We now solve this system of linear equations. First, solve the second equation for \( x \):
[tex]\[ 2 x + 18 y = 0 \implies x = -9 y \][/tex]
Substitute \( x = -9 y \) into the first equation:
[tex]\[ 14 (-9 y) + 2 y + 32 = 0 \][/tex]
[tex]\[ -126 y + 2 y + 32 = 0 \][/tex]
[tex]\[ -124 y + 32 = 0 \][/tex]
[tex]\[ -124 y = -32 \][/tex]
[tex]\[ y = \frac{32}{124} = \frac{8}{31} \][/tex]
Substitute \( y = \frac{8}{31} \) back into \( x = -9 y \):
[tex]\[ x = -9 \left(\frac{8}{31}\right) = -\frac{72}{31} \][/tex]
Therefore, the solution is:
[tex]\[ x = -\frac{72}{31}, \quad y = \frac{8}{31} \][/tex]
Thus, the correct choice is:
[tex]$\square$[/tex] A. There is only one solution where [tex]\( f_x(x, y) = 0 \)[/tex] and [tex]\( f_y(x, y) = 0 \)[/tex], when [tex]\( x = -\frac{72}{31} \)[/tex] and [tex]\( y = \frac{8}{31} \)[/tex].
First, we find the partial derivatives of the function \( f(x, y) \):
1. Compute the partial derivative with respect to \( x \):
[tex]\[ f_x(x, y) = \frac{\partial}{\partial x} (7 x^2 + 9 y^2 + 2 x y + 32 x - 6) \][/tex]
[tex]\[ f_x(x, y) = 14 x + 2 y + 32 \][/tex]
2. Compute the partial derivative with respect to \( y \):
[tex]\[ f_y(x, y) = \frac{\partial}{\partial y} (7 x^2 + 9 y^2 + 2 x y + 32 x - 6) \][/tex]
[tex]\[ f_y(x, y) = 18 y + 2 x \][/tex]
Next, we set the partial derivatives to zero to find the critical points:
[tex]\[ 14 x + 2 y + 32 = 0 \][/tex]
[tex]\[ 18 y + 2 x = 0 \][/tex]
We now solve this system of linear equations. First, solve the second equation for \( x \):
[tex]\[ 2 x + 18 y = 0 \implies x = -9 y \][/tex]
Substitute \( x = -9 y \) into the first equation:
[tex]\[ 14 (-9 y) + 2 y + 32 = 0 \][/tex]
[tex]\[ -126 y + 2 y + 32 = 0 \][/tex]
[tex]\[ -124 y + 32 = 0 \][/tex]
[tex]\[ -124 y = -32 \][/tex]
[tex]\[ y = \frac{32}{124} = \frac{8}{31} \][/tex]
Substitute \( y = \frac{8}{31} \) back into \( x = -9 y \):
[tex]\[ x = -9 \left(\frac{8}{31}\right) = -\frac{72}{31} \][/tex]
Therefore, the solution is:
[tex]\[ x = -\frac{72}{31}, \quad y = \frac{8}{31} \][/tex]
Thus, the correct choice is:
[tex]$\square$[/tex] A. There is only one solution where [tex]\( f_x(x, y) = 0 \)[/tex] and [tex]\( f_y(x, y) = 0 \)[/tex], when [tex]\( x = -\frac{72}{31} \)[/tex] and [tex]\( y = \frac{8}{31} \)[/tex].
