Maximize $\exp(x)+\exp(y)$ over the unit circle $x^2+y^2=1$.

Question

Consider the function $$f(x,y)=\exp(x)+\exp(y).$$ How can I maximize this function over the unit circle $x^2+y^2=1$?

My attempt. I tried using Lagrange multipliers but I was only able to solve the equations numerically an got $x\approx y\approx 0.7$ as the maximum of $f$. Is there a better solution?

Answer 1

We only need to consider $x,y\geq0$. For $0\le x\le 1$, let $$g(x)=\exp(x)+\exp\left(\sqrt{1-x^2}\right).$$

Note that for every $x,y\geq 0$ with $x^2+y^2=1$, we have $f(x,y)=g(x)$. So we want to maximize $g$. Note that $$g'(x)=\exp(x)-\exp\left(\sqrt{1-x^2}\right)\cdot\frac{x}{\sqrt{1-x^2}}.$$

So for $0<x<1$, $$g'(x)=0\iff \exp(x)=\exp(y)\cdot\frac{x}{y}\iff\frac{\exp(x)}x=\frac{\exp(y)}{y},$$

where I have set $y=\sqrt{1-x^2}$.

It is easy to check that $g'\left(\frac{1}{\sqrt2}\right)=0$. Note that the function $x\mapsto \frac{\exp(x)}x$ is strictly monotonically decreasing for $0\le x\le 1$ as can be easily checked by computing the derivative.

Hence, if $x<\frac{1}{\sqrt 2}$, then $y>\frac{1}{\sqrt 2}$ and thus $\frac{\exp(x)}x>\frac{\exp(y)}y$. Same idea if $x>\frac1{\sqrt 2}$.

It follows that $g'(x)=0$ if and only if $x=\frac{1}{\sqrt 2}$.

Now the maximum is simply $\max\left(g(0),g\left(\frac{1}{\sqrt 2}\right),g(1)\right)=\max(1+e,2e^{\frac{1}{\sqrt 2}})=2e^{\frac1{\sqrt{2}}}.$

Answer 2

Use polar coordinates $(r,a)$,

$$f(r,a)=e^{r\cos a}+ e^{r\sin a}, \>\>\>\>\>r\le 1$$

For a given $r$, maximize $f(r,a)$ by setting $f_a’=0$,

$$\tan a= e^{\sqrt2r\sin(a-\frac\pi4)}$$

which yields $a = \pi/4$. Thus, for any $r$, $f(r,a)$ is maximized at $f(r,\frac\pi4) = 2e^{\frac r{\sqrt2}}$, and its overall maximal value is then $f(1,\frac\pi4)=2e^{\frac 1{\sqrt2}}$.

Answer 3

Lagrange works here. You get $e^x = 2\lambda x$ and $e^y = 2\lambda y$. Since $e^x$ and $e^y$ aren't zero, $\lambda$ is nonzero and you have $$xe^{-x} = {1 \over 2\lambda} = ye^{-y} \tag{1}$$ Note that the derivative of $xe^{-x}$ is $(1 - x)e^x$ which is positive on $[-1,1]$ except at $x = 1$ where it is zero. Thus $xe^{-x}$ is monotone increasing on the domain in question, so that $xe^{-x} = ye^{-y}$ from equation $(1)$ implies $x = y$.

Inserting $x = y$ back into $x^2 + y^2 = 1$ gives $(x,y) = ({1 \over \sqrt{2}}, {1 \over \sqrt{2}})$ or $(-{1 \over \sqrt{2}}, -{1 \over \sqrt{2}})$. The first option gives the maximum, namely $2e^{1 \over \sqrt{2}}$.