I think you have a typo in your function. I am assuming you mean 5y, not just 5. OK?.
Your point lies outside the boundary.
\(\displaystyle f(x,y)=1+4x-5y\)....
[1]
Let's try some others that lie on the boundary of the triangle.
The line segment between (0,0) and (2,0):
On this line segment we have y=0, so
[1] simplifies to a function of a single variable x,
f(x,0)=4x+1
This function has no critical points because f'(x)=4 is nonzero for all x.
Thus, the extreme values occur at the endpoints x=0 and x=2, which correspond to (0,0) and (2,0) of D.
The line segment between (0,0) and (0,3):
The same follows with x=0 in our function. The critical values correspond to the endpoints (0,0) and (0,3).
The line segment between (2,0) and (0,3):
The line equation for the line between (2,0) and (0,3) is \(\displaystyle y=\frac{-3x}{2}+3\)
Sub this into
[1]: \(\displaystyle 1+4x-5(\frac{-3x}{2}+3)=\frac{23x}{2}+16\)
This is also nonzero for all values of x, \(\displaystyle f'(x)=\frac{23}{2}\)
Therefore, the critical points occur at (2,0) and (3,0).
Put these points in a chart to see which are the min and the max.
Code:
(x,y)| (0,0) (2,0) (0,3)
----------------------------------------
f(x,y)| 1 9 -14
The max is at (2,0) and the min is at (0,3).
I hope I didn't overlook something. Check it out.
