Question

for the function defined as follows, find all values of x and y such that both ( f_x(x,y)=0 ) and ( f_y(x,y)=0 ).
( f(x,y)=5x^2 + 8y^2 + 4xy + 36x - 6 )
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. there is only one solution where ( f_x(x,y)=0 ) and ( f_y(x,y)=0 ), when ( x=square ) and ( y=square ).
(type integers or simplified fractions.)
b. there are two solutions where ( f_x(x,y)=0 ) and ( f_y(x,y)=0 ), in order from increasing x values, when ( x=square ) and ( y=square ) and ( x=square ) and ( y=square ).
(type integers or simplified fractions.)
c. there are three solutions where ( f_x(x,y)=0 ) and ( f_y(x,y)=0 ), in order from increasing x values, when ( x=square ) and ( y=square ) and ( x=square ) and ( y=square ) and ( x=square ) and ( y=square ).
(type integers or simplified fractions.)
d. there are no solutions where ( f_x(x,y)=0 ) and ( f_y(x,y)=0 ).

Explanation:

Step1: Find the partial derivatives

The function is \(f(x,y) = 5x^{2}+8y^{2}+4xy + 36x-6\).
The partial derivative with respect to \(x\), \(f_{x}(x,y)\):
Using the power rule \((x^{n})^\prime=nx^{n - 1}\) and \((uv)^\prime = u^\prime v+uv^\prime\) (where \(u = 4y\) and \(v=x\) for the \(4xy\) term), we have \(f_{x}(x,y)=10x + 4y+36\).
The partial derivative with respect to \(y\), \(f_{y}(x,y)\):
\(f_{y}(x,y)=16y + 4x\).

Step2: Set up the system of equations

Set \(f_{x}(x,y)=0\) and \(f_{y}(x,y)=0\).
We get the system of equations:
\(

$$\begin{cases}10x + 4y+36=0\\4x + 16y=0\end{cases}$$

\)
From the second equation \(4x+16y = 0\), we can simplify it to \(x=- 4y\).
Substitute \(x = - 4y\) into the first equation:
\(10(-4y)+4y+36=0\).
Expand: \(-40y+4y + 36=0\).
Combine like - terms: \(-36y+36 = 0\).
Solve for \(y\): \(-36y=-36\), so \(y = 1\).
Substitute \(y = 1\) into \(x=-4y\), then \(x=-4\).

Answer:

A. There is only one solution where \(f_{x}(x,y)=0\) and \(f_{y}(x,y)=0\), when \(x=-4\) and \(y = 1\).