If $x>0,y>0$ and $4xy=2^{x+y}$ then find the minimum and maximum values of $x+y$.

Question

If $x>0,y>0$ and $4xy=2^{x+y}$

then find the minimum and maximum values of $x+y$.

My Attempt

I tried by putting $t=x+y$

$\Rightarrow 4x(t-x)=2^t$.

On differentiation we have $4t-8x=\frac{dt}{dx}(2^tln2-4x)$.

For maximum/minimum put $\frac{dt}{dx}=0$ to obtain $t=2x$

$\Rightarrow x+y=2x$ and thus $y=x$.

Putting in given equation one obtains $4x^2=2^{2x}$.

The obvious solution here is $x=1,2$.

So the value of corresponding $y$ will be $y=1,2$.

So, the extreme values of $x+y$ can be $2$ and $4$.

Is above correct or am I missing something here.

Can we solve this using $AM\geq GM$ inequality.

Answer 1

By AM-GM, we have $$(x + y)^2 \ge 4xy = 2^{x + y} \tag{1}$$ which results in $2 \le x + y \le 4$.
(Note: Letting $u := x + y > 0$, from (1), we have $u \ge 2^{u/2}$. Using calculus, we can prove that $2 \le u \le 4$.)

Since $x = y = 2$ is feasible, the maximum is $4$. Since $x = y = 1$ is feasible, the minimum is $2$.

Answer 2

Let $x$ and $y$ be positive real numbers such that $4xy=2^{x+y}$. Set $t:=x+y$ so that $xy=2^{t-2}$ and hence $x$ and $y$ are real roots of the quadratic $$X^2-tX+2^{t-2}.$$ This implies that the discriminant $\Delta$ of this quadratic is nonnegative, where $$\Delta=(-t)^2-4\cdot2^{t-2}=t^2-2^t,$$ and it follows that $2\leq t\leq 4$. For $t=2$ we find $x=y=1$ and for $t=4$ we find $x=y=2$.