# Min max problem

## Homework Statement ## Homework Equations

marginal revenue[/B] (R') is the additional revenue that will be generated by increasing product sales by one unit

## The Attempt at a Solution

I don't know how to start. Q is the number of items sold at price x. y is the marginal cost, the cost of producing one item. N is the net profit, the revenue (my english isn't perfect so i explain each variable) $~N=Q(x-y)~$.
The derivative N' is the marginal revenue, no?
Q, x and y are variables. x and y can change according to Q, because if i produce more i can lower the price i charge (x) and also the cost y.
How do i express marginal revenue?

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Quantity sold/quantity produced $Q=Q(P)$, meaning $Q$ is a function of $P$, where $Q$ is the quantity sold at price $P$. $\\$ Revenue $R=Q \cdot P$. $\\$ The cost to make $Q$ items is a function of $Q$, so that $C=C(Q)$. $\\$ Net profit $N=R-C$. $\\$ Since $Q=Q(P)$, we can also write it in the form $P=P(Q)$, i.e. $P$ is a function of $Q$. Thereby, revenue $R=Q \cdot P=Q \cdot P(Q)$ is a function of $Q$, so we can write $R=R(Q)$. $\\$ We see that $N=N(Q)=R(Q)-C(Q)$. $\\$ How do we maximize $N$, using calculus? Also, what is the definition of marginal revenue, using calculus? And what is the definition of marginal cost, using calculus? $\\$ Once we find the $Q$ where marginal revenue =marginal cost, we can then go to the function $P=P(Q)$, the inverse function of $Q=Q(P)$, to determine what price we should charge. The net profit $N$ will be maximized at that selling price.

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• Delta2
Marginal revenue=R':
$$R=Q \cdot P=Q \cdot P(Q)~\rightarrow~R'=P(Q)+Q\cdot P'(Q)$$
Marginal cost=C': $~C=C(Q)~\rightarrow~C'=C'(Q)$
Marginal revenue=marginal cost: $~P(Q)+Q\cdot P'(Q)=C'(Q)$
It doesn't help. C=C(Q) needn't be $~C=Q\cdot k~$, the more items you make it's cheaper for each item.
And also, i have to prove that when Marginal revenue=marginal cost N is biggest

The maximum or minimum in $N$ occurs when $\frac{dN}{dQ}=0$. Buy the above, this clearly is when $\frac{dR}{dQ}-\frac{dC}{dQ}= 0$ so that $\frac{dR}{dQ}=\frac{dC}{dQ}$. The marginal revenue calculus definition is $\frac{dR}{dQ}$. Similarly, the marginal cost calculus definition is $\frac{dC}{dQ}$. I basically gave you the complete proof... $\\$ I have not proven that it is indeed a maximum. If you can show $\frac{d^2 N}{dQ^2} <0$, then it is a maximum.
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