## Calculus 8th Edition

Absolute minimum: $f(1,1)=f(-1,1)=7$ or, $f(\pm 1,1)=7$ Absolute maximum: $f(0,0)=4$
To calculate the critical points we have to first put $f_x(x,y), f_y(x,y)$ equal to $0$. Thus, $f_x=2x+2xy,f_y=2y+x^2$ This yields, $x=0,y=0$ and $f(1,1)=1$ From the vertices of the triangle, after simplifications, we have At $x=-1$; $f(y)=1+y^2+y+4=y^2+y+5$ and $f'(y)=2y+1$ Thus, Critical point: $(-1,-\dfrac{1}{2})$ At $x=1$; $f(y)=1+y^2+y+4=y^2+y+5$ and $f'(y)=2y+1$ Thus, Critical point: $(1,-\dfrac{1}{2})$ At $y=-1$; $f(x)=1-x^2+x+4=5$ Thus, No Critical point At $y=-1$; $f(x)=x^2+1+x^2+4=2x^2+5$ and $f'(x)=4x$ Thus, Critical point: $(0,1)$ Therefore, it has minimum and maximum value are at $x=7,4$ Hence, Absolute minimum: $f(1,1)=f(-1,1)=7$ or, $f(\pm 1,1)=7$ Absolute maximum: $f(0,0)=4$