I need to find the absolute minimum of
$$ f(x,y)=x^2+5xy+3y^2-4x-2y-5 $$
on the domain $$-3\le x\le9, 1\le y\le10$$
I have attempted to find the boundaries with
$$f(-3,y)=3y^2-17y+16$$$$f'(-3,y)=6y-17=0$$$$y=17/6$$$$f(-3,17/6)=-225/4$$
but it seems this is not the correct answer. Where am I going wrong?
Edit: Trying to find critical points in the interior $$2x+5y-4=0$$$$6y+5x-2=0$$$$x=-14/13,y=16/13$$$$f(-14/13,16/13)=-2225/169$$which is still incorrect.
Trying with x=9 $$f(9,y)=43y+3y^2+40$$$$f'(9,y)=6y+43=0$$$$y=-43/6$$ which I am confused about why this does not fit in the constraint inequality?
Calculating f(x,1) I end up with $$f(x,y)=-41/4$$ For f(x,10) I get -23 which again does not work with the constraint.
The corners are as follows:$$f(-3,1)=-4$$$$f(-3,10)=-454$$$$f(9,1)=80$$$$f(9,10)=170$$
however none of these are working for me as the correct minimum?