Answered

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For the function defined as follows, find all values of [tex]$x$[/tex] and [tex]$y$[/tex] such that both [tex]$f_x(x, y)=0$[/tex] and [tex]$f_y(x, y)=0$[/tex].

[tex]\[ f(x, y) = 7x^2 + 9y^2 + 2xy + 32x - 6 \][/tex]

Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

A. There is only one solution where [tex]$f_x(x, y)=0$[/tex] and [tex]$f_y(x, y)=0$[/tex], when [tex]$x =\square$[/tex] and [tex]$y =\square$[/tex]. (Type integers or simplified fractions.)

B. There are two solutions where [tex]$f_x(x, y)=0$[/tex] and [tex]$f_y(x, y)=0$[/tex], in order from increasing [tex]$x$[/tex] values, when [tex]$x=$[/tex] [tex]$\square$[/tex] and [tex]$y=$[/tex] [tex]$\square$[/tex] and [tex]$x=$[/tex] [tex]$\square$[/tex] and [tex]$y=$[/tex] [tex]$\square$[/tex]. (Type integers or simplified fractions.)

C. There are three solutions where [tex]$f_x(x, y)=0$[/tex] and [tex]$f_y(x, y)=0$[/tex], in order from increasing [tex]$x$[/tex] values, when [tex]$x=$[/tex] [tex]$\square$[/tex] and [tex]$y =$[/tex] [tex]$\square$[/tex] and [tex]$x =$[/tex] [tex]$\square$[/tex] and [tex]$y =$[/tex] [tex]$\square$[/tex] and [tex]$x =$[/tex] [tex]$\square$[/tex]. (Type integers or simplified fractions.)

D. There are no solutions where [tex]$f_x(x, y)=0$[/tex] and [tex]$f_y(x, y)=0$[/tex].

Sagot :

GinnyAnswer
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].
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