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I am trying to find the values $x \in [0, 16]$ and $y \in [0, 12]$ that maximise

$$ f(x,y) \equiv \min \{\ln(x) + 2\ln(y), \ln(16-x) + 2\ln(12-y) \}$$

Intuitively, it is pretty clear that the solution is $x = 8$, $y = 6$. However, I am having a bit of trouble proving this. Can anyone give me a hint?

My thoughts so far are, in order to maximise the minimum of the functions, we should equate the two arguments of the min function. Otherwise, we could increase the smaller argument and decrease the larger argument, thus increasing $f$. So we get the equation

$$ \ln(x) + 2\ln(y) = \ln(16-x) + 2\ln(12-y)$$

Now one solution to this equation is $x = 16 - x$, $y = 12 - y$, yielding $x = 8$ and $y = 6$. However, this is not the only solution; and it's not obvious to me how to show that the solution I found is the optimum.

afreelunch
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1 Answers1

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Because $\ln x$ is monotonic over the reals, the curve of intersection can be described by

$$xy^2= (16-x)(y-12)^2$$

It can be shown that $x = 0$, $y=12$ does not produce the maximum, so we can isolate for $x$ as follows

$$\frac{y^2}{(y-12)^2} = \frac{16-x}{x} = \frac{16}{x} - 1$$ $$\frac{2y^2 - 24y + 144}{(y-12)^2} = \frac{16}{x}$$

As $2y^2-24y+144 > 0 \; \forall x$ $$x = \frac{8(y-12)^2}{y^2-12y+72}$$

Then it's about maximizing

$$f(y) = \frac{8(y-12)^2y^2}{y^2-12y+72}$$

Once we have established that the only critical value on $y \in (0, 12)$ is $y=6$, and that $f(0)=f(12) < f(6)$, we obtain that the maximum is located at $x=8, y=6$.

okzoomer
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