extremum under constraint question I was tasked with finding the extremum of $z=xy$ under the constraint $x+y=1$, here is what I did: $$z=xy$$ $$x+y=1$$ from the second line we get $y=x-1$ and we substitute that back in the first equation: $$z=x(1-x)= x-x^2$$
$$z'(x) = 1-2x$$ next $$z'(x) = 0$$ $$&lt;=&gt; $$ $$0 = 1-2x$$ $$x=0.5$$ $$=&gt;y=0.5$$ $$=&gt;$$ $$(0.5,0.5)$$ suspected as extrema. 
$$z''(x)= -2$$ $$z''(0.5)= -2&lt;0 $$ hence $$(0.5,0.5)$$ local maxima. $$=&gt; z_{max} = z(0.5,0.5) = 0.5 * 0.5 = 0.25 $$

Is this ok?
hint: $(x-y)^2 \geq 0 \implies xy \leq \dfrac{(x+y)^2}{4}$. Can you take it from here?
Simple problem with a nice graph: