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)=2x - 1.6 )

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

Explanation:

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

Differentiate \(f(x,y)=4.1 - 1.6x+7.3y + 0.3y^{3}+x^{2}\) with respect to \(y\).
Using the power rule \(\frac{d}{dy}(y^{n})=ny^{n - 1}\), and treating \(x\) as a constant.
The derivative of \(4.1\) with respect to \(y\) is \(0\) (since it's a constant), the derivative of \(-1.6x\) with respect to \(y\) is \(0\) (since \(x\) is treated as a constant), the derivative of \(7.3y\) with respect to \(y\) is \(7.3\), the derivative of \(0.3y^{3}\) with respect to \(y\) is \(0.3\times3y^{2}=0.9y^{2}\), and the derivative of \(x^{2}\) with respect to \(y\) is \(0\) (since \(x\) is treated as a constant).
So, \(f_y(x,y)=7.3 + 0.9y^{2}\)

Step2: Analyze the critical - points

To find the critical points, we set \(f_x(x,y)=0\) and \(f_y(x,y)=0\).
Set \(f_x(x,y)=2x - 1.6=0\), then \(2x=1.6\), and \(x = 0.8\).
Set \(f_y(x,y)=7.3+0.9y^{2}=0\).
We have \(0.9y^{2}=-7.3\), or \(y^{2}=-\frac{7.3}{0.9}\approx - 8.11\).
Since \(y^{2}\geq0\) for all real \(y\), the equation \(f_y(x,y) = 0\) has no real - valued solutions.

Answer:

\(f_y(x,y)=7.3 + 0.9y^{2}\). Since the system of equations \(f_x(x,y)=0\) and \(f_y(x,y)=0\) has no solution (because \(f_y(x,y)=7.3 + 0.9y^{2}=0\) has no real \(y\) - values), the function \(f(x,y)\) has no critical points. And since local extrema occur at critical points (points where \(f_x(x,y)=0\) and \(f_y(x,y)=0\)), the function \(f(x,y)\) has no local extrema.