Question

find $f_x(x,y)$ and $f_y(x,y)$, and explain why $f(x,y)$ has no local extrema.

$f(x,y)=4.1 - 1.6x + 7.3y + 0.3y^3 + x^2$

$f_x(x,y)=\square$

Explanation:

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

Differentiate \(f(x,y)=4.1 - 1.6x+7.3y + 0.3y^{3}+x^{2}\) with respect to \(x\) (treating \(y\) as a constant).
Using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\) and \(\frac{d}{dx}(c)=0\) (where \(c\) is a constant), we have:
\(f_x(x,y)=\frac{\partial}{\partial x}(4.1)-\frac{\partial}{\partial x}(1.6x)+\frac{\partial}{\partial x}(7.3y)+\frac{\partial}{\partial x}(0.3y^{3})+\frac{\partial}{\partial x}(x^{2})\)
\(f_x(x,y)=0 - 1.6+0 + 0+2x=2x - 1.6\)

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

Differentiate \(f(x,y)=4.1 - 1.6x+7.3y + 0.3y^{3}+x^{2}\) with respect to \(y\) (treating \(x\) as a constant).
Using the power rule \(\frac{d}{dy}(y^{n})=ny^{n - 1}\) and \(\frac{d}{dy}(c)=0\) (where \(c\) is a constant), we have:
\(f_y(x,y)=\frac{\partial}{\partial y}(4.1)-\frac{\partial}{\partial y}(1.6x)+\frac{\partial}{\partial y}(7.3y)+\frac{\partial}{\partial y}(0.3y^{3})+\frac{\partial}{\partial y}(x^{2})\)
\(f_y(x,y)=0-0 + 7.3+0.9y^{2}+0=0.9y^{2}+7.3\)

Step3: Analyze for local extrema

For a function \(z = f(x,y)\) to have a local extremum, we need to find critical points by setting \(f_x(x,y)=0\) and \(f_y(x,y)=0\).
Set \(f_x(x,y)=0\): \(2x-1.6 = 0\Rightarrow x = 0.8\)
Set \(f_y(x,y)=0\): \(0.9y^{2}+7.3=0\Rightarrow y^{2}=-\frac{7.3}{0.9}\approx - 8.11\)
Since \(y^{2}\geq0\) for all real \(y\), the equation \(f_y(x,y) = 0\) has no real - valued solutions.

Answer:

\(f_x(x,y)=2x - 1.6\)