I have been trying to solve the following function:
$$Q(x,y) = \exp (0.5077 - 0.324 \ln(x))x \cdot y$$
My interest is to find the value of $ x $ which maximizes this function. Therefore I proceed by taking $\frac{\partial Q}{\partial x} = 0$ but can’t go beyond this point as I get
$$\dfrac{\partial Q}{\partial x} = \dfrac{1.12315 y}{x^{0.324}} = 0 $$
I am wondering if a numerical solution instead exists for this problem.
*Clarification:
For simplicity assume $ Q(x,y) = \exp(a + b \ln x) x \cdot y $ where $a = 0.5077$ and $b = -0.324$ as clarified by caverac below. Then I want to take $ \frac{\partial Q}{\partial x} = 0$ and get the value of $x$ which maximises $Q(x,y)$.
For simplicity call $a = 0.5077$ and $b = -0.324$, so that you can write your problem as
$$ Q(x,y) = \exp(a + b\ln x) xy = e^a e^{b\ln x} xy = e^a x^bxy = e^a x^{1 + b}y $$
So the problem is reduced to finding the optimum value of
$$ f(x,y) = x^{1 + b}y = x^{0.676}y $$
$(0, 0)$ is a saddle point, that means it is not a bound problem