Solution:
There is no saddle point (DNE). However, there is local maximum at (1, 1/2) for the given function.
Explanation:
we have function of two variables f(x,y)= 9-2x+4y-x^2-4y^2
we will find the values by partial derivative with respect to x,y,xy
[tex]f_{x}[/tex]= -2 -2x
[tex]f_{y}[/tex]= 4 -8y
to find the saddle point we should first find the critical points so equate
-2 -2x=0 and 4 -8y=0
we get x= 1 and y =1/2 so, critical points are (1,1/2)
to find local maximum or minimum we have to find [tex]f_{xx}[/tex], [tex]f_{yy}[/tex] and [tex]f_{xy}[/tex]
formula is [tex]f_{xx}[/tex] *[tex]f_{yy}[/tex] - [tex]f^{2_{xy} }[/tex] =0
[tex]f_{xx}[/tex] = -2
[tex]f_{yy}[/tex] = -8
[tex]f^{2_{xy} }[/tex] =0
putting values in formula
(-2)*(-8) -0 =16 > 0, and [tex]f_{xx}[/tex]< 0 and [tex]f_{yy}[/tex]<0
so, here we have local maximum
we have no saddle point for this function by using the same formula we used to find extrema.
Answer:
The point (-1,1/2) is a local maximum and there are no local minimums and saddle points.
Step-by-step explanation:
The function to study is [tex]f(x,y) = 9-2x+4y-x^2-4y^2[/tex] which is a differentiable function on the variables [tex]x[/tex] and [tex]y[/tex]. We find the possible points of local maximum, local minimum and saddle points finding the point where the partial derivatives are zero at the same time. So, we must calculate the partial derivatives:
[tex]\frac{\partial f}{\partial x} = -2-2x = -2(x+1)[/tex]
[tex]\frac{\partial f}{\partial y} = 4-8y = 4(1-2y)[/tex]
It is not difficult to see that the only point where [tex]\frac{\partial f}{\partial x}(x,y)=\frac{\partial f}{\partial y}(x,y) =0[/tex] is (-1,1/2). Notice that the system of equations
[tex]\begin{cases} -2(x+1) &= 0\\ 4(1-2y) &= 0\end{cases}[/tex]
has solution x= -1 and y=1/2.
If we want to know if (-1,1/2) is local maximum, local minimum or saddle point we must calculate the partial derivatives of second order:
[tex]\frac{\partial^2 f}{\partial x^2} = -2 = A[/tex]
[tex]\frac{\partial^2 f}{\partial y^2} = -8 = B[/tex]
[tex]\frac{\partial^2 f}{\partial y \partial x} = \frac{\partial^2 f}{\partial x \partial y} = 0 = C[/tex].
Substituting the values into the formula [tex]AB-C^2[/tex] we get (-2)(-8)-0=16>0. As the value we have obtained is greater than zero and A=-2<0 we assure that the point (-1,1/2) is a local minimum.
There are no local maximums or saddle points.
