Firstly, I find the critical points:
$f_x=8y=0$; $f_y=8x+1=0$
From here, $y=0$ and $x= -{{1} \over 8}$ and I find the point $P(0, -{{1} \over 8})$ which, does not satisfy the condition under $x$, therefore, there are no critical points in the interior of the region. (<- someone confirm this).
Because there is no critical point, my next step is to evaluate at endpoints:
$B_1$: $y=0$, $0≤x≤5$, which follows that $f(x,0)=0$ for any $x$, so that means that at point $(0,0)$, the function has a saddle (can confirm from a 3d graph), but then again, (5,0) would also mean there is a saddle or a minima. How do I confirm this mathematically?
$B_2$: $x=5$, $0≤y≤10$, which follows that $f(5,y)=40y+y$. If I find the derivate from $f_y(5,y)=40+1 \neq 0$, it shows that it is non-existent, which would mean there are no extrema in that part. (<- someone confirm this).
$B_3$: $0≤x≤5$, $10≤y≤15-x$, which follows that $f(x,y)=8xy+y$. By logic, I should only search the part at the point $(x=0,15-x) = (0,15)$, which yields $f(0,15)=8*0*15+15=15$. This would be the first maxima. (<- someone confirm this).
$B_4$: $x=0$, $0≤y≤15$ which follows that $f(0,y)=y$. For $(0,0)$ I get the saddle point again, and for $(0,15)$ I would get $f(0,15)=8*0*15+15=15$ again, which would only confirm my maxima.
What I really need is, someone to confirm if my calculations are true.