Question

find ( f_{x}(x,y) ) and ( f_{y}(x,y) ), and explain why ( f(x,y) ) has no local extrema.

( f(x,y)=5x + 5y + 7 )

( f_{x}(x,y)=5 )

( f_{y}(x,y)=square )

Explanation:

Step1: Find \( f_y(x,y) \)

Differentiate \( f(x,y)=5x + 5y+7 \) with respect to \( y \).
Using the rule that if \( z = ax+by + c\) (\(a,b,c\) are constants), then \(\frac{\partial z}{\partial y}=b\).
So \( f_y(x,y)=5 \).

Step2: Analyze critical points

For a function \( z = f(x,y) \), critical points occur where \( f_x(x,y)=0 \) and \( f_y(x,y)=0 \).
Here \( f_x(x,y) = 5
eq0\) and \( f_y(x,y)=5
eq0 \).
Since there are no points \((x,y)\) such that \( f_x(x,y)=0\) and \( f_y(x,y)=0 \), there are no critical points.
By the second - derivative test (which requires critical points \((x_0,y_0)\) where \( f_x(x_0,y_0)=f_y(x_0,y_0) = 0\) to start with), a function \( f(x,y)\) can have local extrema only at critical points.

Answer:

\( f_y(x,y)=5 \). The function \( f(x,y) = 5x+5y + 7\) has no local extrema because there are no points \((x,y)\) such that \( f_x(x,y)=0\) and \( f_y(x,y)=0\) (no critical points exist).