Here are some solutions from Koopa,Boston College, USA. Vassil, Lawnswood Sixth Form, Leeds sent in similar methods. Can you find a co-ordinate geometry (i.e. graphical) method or yet another different method?

Method 1 We have $x+y=-1$. So, to maximise $\mathrm{xy}$, I need to maximize

${\displaystyle -x(x+1)=-[(x+1/2{)}^2-1/4]=-(x+1/2{)}^2+1/4,}$

so, $\mathrm{xy}$ is maximized at x = -1/2 and the maximum value is 1/4.

Method 2 Let $f(x)=-x(x+1)$, then by differentiation $f\text{'}(x)=-2x-1$ and to find a maximum or minimum $f\text{'}(x)=0$ gives $x=-1/2$. The second derivative test easi ly verifies that this indeed gives a maximum so the maximum value is 1/4.

Method 3 By the AM-GM inequality, we have

${\displaystyle (\mathrm{xy}{)}^{(1/2)}\le (x+y)/2}$

so

${\displaystyle \mathrm{xy}\le (-1/2{)}^2=1/4.}$