Question

find $f_x(x,y)$ and $f_y(x,y)$, and explain why $f(x,y)$ has no local extrema.
$f(x,y)=5x + 5y + 7$
$f_x(x,y)=\square$

Explanation:

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

Differentiate \( f(x,y)=5x + 5y+7 \) with respect to \( x \) (treating \( y \) as a constant).
Using the power rule \( \frac{\partial}{\partial x}(ax^n)=nax^{n - 1} \), for \( n = 1 \) and \( a=5 \) in the term \( 5x \), and \( \frac{\partial}{\partial x}(c)=0 \) (where \( c = 5y+7 \) is a constant with respect to \( x \)).
\( f_x(x,y)=\frac{\partial}{\partial x}(5x)+\frac{\partial}{\partial x}(5y)+\frac{\partial}{\partial x}(7)=5+0 + 0=5 \)

Step2: Find \( f_y(x,y) \)

Differentiate \( f(x,y)=5x + 5y+7 \) with respect to \( y \) (treating \( x \) as a constant).
Using the power rule \( \frac{\partial}{\partial y}(ay^n)=nay^{n - 1} \), for \( n = 1 \) and \( a = 5 \) in the term \( 5y \), and \( \frac{\partial}{\partial y}(c)=0 \) (where \( c=5x + 7 \) is a constant with respect to \( y \)).
\( f_y(x,y)=\frac{\partial}{\partial y}(5x)+\frac{\partial}{\partial y}(5y)+\frac{\partial}{\partial y}(7)=0+5 + 0=5 \)

Step3: Analyze local extrema

For a function \( z = f(x,y) \) to have a local extremum, we need to find critical points. Critical points occur where \( f_x(x,y)=0 \) and \( f_y(x,y)=0 \).
Since \( f_x(x,y)=5
eq0 \) and \( f_y(x,y)=5
eq0 \) for all \( (x,y)\in\mathbb{R}^2 \), there are no points \( (x_0,y_0) \) such that \( f_x(x_0,y_0)=0 \) and \( f_y(x_0,y_0)=0 \). So, the function \( f(x,y) \) has no critical points. And since local extrema (local maxima or minima) can only occur at critical points, \( f(x,y) \) has no local extrema.

Answer:

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