How do I find the maximum value of $x^{\ln y}$?

Question

Let $$x\gt1, y\gt1$$ and $$(\ln x)^2+(\ln y)^2=\ln x^2+\ln y^2$$

Find the maximum value of $$x^{\ln y}$$

Where do I start and where do I end?

Answer 1

Let: $$\ln x = a~ , ~~\ln y =b ~, ~~M=x^{lny}$$

$$M=x^{\ln y}=e^{\ln x \ln y}$$

Since $e^t$ is strictly increasing, when $\ln x \times \ln y$ will be maximum, $M$ will be maximum.

Given :$$a^2+b^2=2a+2b ~~~~~~\dots (1)$$ We want to maximize product $ab$.

Using $\text{A.M.} \ge\text{ G.M}$ for $a^2$ and $b^2$ :

$$\frac{a^2+b^2}{2} \ge \sqrt{a^2\times b^2}$$

$$a+b \ge ab $$

Equality occurs when $a=b$.Put in $(1)$

On solving; $a=0,2$

Since $x,y >1$, $a=0~~ \text{rejected}$

Therefore :

$$ \max{\text{M}} = e^{ab}=e^4$$

Answer 2

Let $\ln x=a$ and $\ln y=b$.

Hence, $a>0$, $b>0$ and $a^2+b^2=2(a+b)$. Thus, by AM-GM $$2(a+b)=a^2+b^2\geq\frac{(a+b)^2}{2},$$ which gives $a+b\leq4$. Hence, by AM-GM again $4\geq a+b\geq2\sqrt{ab}$, which gives $ab\leq4$.

Thus, $x^{\ln{y}}=e^{\ln{x}\ln{y}}=e^{ab}\leq e^4$.

The equality occurs for $a=b=2$ or $x=y=e^2$, which gives the answer: $e^4$.

Answer 3

Denote $\ln{x}$ by $a$, $\ln{y}$ by $b$. We know

$$a, b >0,$$ $$a^2 + b^2 = 2a + 2b,$$ $$x^{\ln{y}}\to\max\Rightarrow ab\to\max.$$

We can now draw all such points $(a, b)$ on a coordinate plane and see that $ab$ is maximal when $a=b=2$:

But, of course, we can do better. For example, we can see that $ab$ has a maximum if we allow $a$ or $b$ equal zero (and still require that $(a-1)^2+(b-1)^2=2$). Let $(a_0, b_0)$ be such point. Obviously, $(b_0, a_0)$ has the same value. Then $((a_0 + b_0)/2, (a_0 + b_0)/2)$ has a greater product of coordinates, but for it $(a - 1)^2 + (b - 1)^2 < 2$, however, $a$ and $b$ exceed one. So we can shift it by some vector $(t, t)$ so that it'll give us a better maximal value, so it has to be reached in $(2, 2)$.