Functions of Several Variables — Question 7

Question 7 – Solution

Strategy Complete squares in a way that respects the mixed term.

Rewrite Since $2x^2+2xy+y^2=(x+y)^2+x^2$, $$f=7-(x+y)^2-x^2+8x+6y.$$ Set $u=x+y$. Then $y=u-x$ and $f=7-u^2-x^2+2x+6u=17-(x-1)^2-(u-3)^2$.

Conclusion Therefore $$\boxed{f(x,y)=17-(x-1)^2-(x+y-3)^2}.$$ The global maximum is $17$ at $x=1$, $x+y=3$, hence $(1,2)$. There is no minimum, and the range is $(-\infty,17]$.