How to maximize a minimum of two functions I would like to maximize $\min_{(x,y)\in(0,1]\times(0,1]}(\frac{4x}{x+y},\frac{6y}{x+2y})$. I am thinking about first considering lines $y=cx$ for some $c\ge0$, so the problem becomes $\max\min_{c\ge0}(\frac{4}{1+c},\frac{6c}{1+2c})$. If $c^*$ is the optimal value for this second problem, I can check on the line $y=c^*x$ to see when the first problem is maximized.
Plotting the second problem, it looks like the optimal value is given when $4/(1+c)=6c/(1+2c)$, i.e. $c=1$. So the optimal value for the first problem is given by $(\sqrt 2, \sqrt 2)$.
I don't think I've missed any values of $(x,y)$ in the domain I'm interested in by considering only rays from the origin, but I'd like to confirm that this is the case.
Also, is there any way to do this problem without resorting to plotting?
Thanks!
 A: The general method for doing this without resporting to plotting is to find the regions where each of the expressions is the minimal one.  For instance, in this problem,  \begin{align*}
    \frac{4x}{x+y} &< \frac{6y}{x+2y}  \\
    (4x)(x+2y) &< (6y)(x+y)  &  &\text{comment 1}  \\
    4x^2 + 8xy &< 6y^2 + 6xy  \\
    2(x-y)(2x+3y) &< 0  &  &\text{comment 2} \\
    x &< y
\end{align*}


*

*comment 1: We have multiplied our inequality by a pair of expressions.  For each that is negative, we must reverse the sense of the inequality.  If either expression can be zero, we must break out that case separately (otherwise, we have written $a < b \implies 0 < 0$, which is clearly false).  Both variables are constrained to be positive, so we have multiplied by expressions which are always positive, and both expressions are additions, so we have not multiplied by zero.

*comment 2:  Since both $2 > 0$ and $2x+3y > 0$ always on our domain of interest, it must be that $(x-y)<0$.


If, for completeness, we check the other inequality, we get $y > x$.  Consequently in each triangular half of the square, split along the line $y=x$, we know which of the two expressions is least.  So we use normal techniques on each triangle to find the maximum of the one function of interest on that region.  We should also check all the boundaries, $x \in\{0,1\}$, $y \in \{0,1\}$, and $x=y$ for their maxima.  (These are "easier" in the sense that they are one dimension smaller.  One may still need to break them up into regions, as above, to find the maxima on each, especially along the line $y=x$ since, strictly, we have not shown which expressions is least along that line segment.)
A: I would start by considering where the two functions are equal to each other.
$\frac{4x}{x+y}=\frac{6y}{x+2y}$
$4x(x+2y)=6y(x+y)$
$4x^2+8xy=6xy+6y^2$
$4x^2+2xy-6y^2=0$
$2x^2+xy-3y^2=0$
$x=\frac{-y \pm \sqrt{y^2-4.2.(-3y^2)}}4$
$x=\frac{-y \pm 5y}4$
$x=-\frac 32y$ or $x=y$
You're only considering the first quadrant, so all that matters is $x=y$. Along this line the two functions are equal in size, above it one function is always smaller and below it the other function is always smaller.
On the line $y=x$ both functions take the value $\frac {4x}{x+y}=\frac {4x}{x+x}=\frac {4x}{2x}=2$
To determine which function is smaller in the region $y>x$ use the test point $(\frac 12, 1)$
$\frac{4x}{x+y}=\frac{2}{1.5}=\frac{4}{3}$
$\frac{6y}{x+2y}=\frac{6}{2.5}=\frac{12}{5}$
So for this region $\frac{4x}{x+y}$ is the smaller function; use calculus to determine where this is maximised.
For the other region $\frac{6y}{x+2y}$ is the smaller function; use calculus to determine where this is maximised.
