Find the highest point on curve defined by intersection of the graph of $g(x,y) = \sqrt{xy}$ and plane $x+y-1=0$

Question

So far this is what I have done:

$$F(x,y) = \sqrt{xy} + λ(x+y-1) =0$$

$$F_x = \frac12(xy)^\left(\frac{-1}{2}\right).y + λ=0$$

$$F_y = \frac12(xy)^\left(\frac{-1}{2}\right).x + λ=0$$

$$Fλ = x+y-1=0$$

I found that $x=y$ and $x=y=\frac12$

I'm not sure if this makes sense and where to go from there.

Answer 1

This $3$D problem can be brought back to a $2$D issue:

From $x+y=1$, one has $y=1-x$. Then we have to find the maximum of $z=\sqrt{x(1-x)}$ which occurs at $x=\frac{1}{2}$ with value $z=\frac{1}{2}$ $\left(\text{using derivatives or the fact that the maximum value for the product of quantities whose sum is constant, here}\; x\; \text{and}\; 1-x, \text{is reached when they are equal}\; x=1-x \Rightarrow x=\frac12\right)$

Remark: In geometrical terms, we have sliced the surface with equation $z=\sqrt{xy}$ with vertical plane $x+y=1$, and have had a reasoning on the intersection curve (a circle).

Answer 2

If you use Lagrange's method in order to find a global extremum you should be aware that this method only brings conditionally stationary points of the objective function $f$ in the relative interior of the considered intersection curve $\gamma$ to the fore. This implies that the argument then has to be supplemented with some global considerations: Do we really have a maximum at this point? What happens at the boundary? Etc.

In the example at hand it suffices to use the AM-GM-inequality: We may assume $x\geq0$, $y\geq0$ and then have $$\sqrt{xy}\leq{x+y\over2}={1\over2}$$ with equality sign iff $x=y={1\over2}$. It follows that the "highest" point on $\gamma$ is the point $\bigl({1\over2},{1\over2},{1\over2}\bigr)$.