It costs a company $\$2$ per unit to make its product. If $A$ dollars are spent per month on advertising, then the number of units per month that will be sold is given by $$ x=30(1-e^{-0.001A})(22-p) $$ where $p$ is the selling price. Find the values of $A$ and $p$ that will maximize the firm's net monthly profit and calculate the value of this maximum profit.
it is known that the profit is given by $$ U= (px)-(2x+A) $$ and by partially deriving and setting to zero, i have
$\frac{\partial U}{\partial p}=30(1-e^{-0.001A})(-2)(p-12)=0$, then $A=0$ and $p=12$.
$\frac{\partial U}{\partial A}=0.03(22−p)(p−2)e^{-0.001A}-1=0$, then $p=2$ and $p=22$, but $A$ is not defined.
Also, there must always be an objective function, and a function to replace, I have no idea which is the function to replace. I really appreciate the help
The function is not differentiable when $A=0$, since you can't spend negative dollars on advertising. So, your analysis of the first equation gives $p=12$. Now substitute $p=12$ in the second equation, and solve for $A$. Both conditions must be satisfied at a critical point, so you can the result from the first equation to the second.
You still have to consider what happens when $A=0$, but the original equation shows that if $A=0$, then $x=0$ and this is not the maximum, I hope.
Perhaps I should add that I didn't check the differentiation.