I need help

Profit is the difference between cost to sell something and income for selling something.
* Given those somethings are sold for $7/unit, income = $7 × unitsSold
* A look-up table has been given for cost as a function of units sold; cost = lookUpCost(unitsSold)

profit = income - cost
profit = $7 × unitsSold - LookUpCost(unitsSold)
Then find the maximum profit for unitsSold = [0...7]

If we created a formula estimating cost as a function of units sold instead of a look-up table, then that iteration over all possibilities would go away.

Hi.

Firstly, as you can notice, the TC function is a parabola (try and draw the graphics if you don't believe me).
Therefore, you can get the function of the parabola by setting a system of equations.

The total cost will be like this: TC = aQ^2+ bQ+c, where Q is the quantity produced.
• 10 = a (0)^2+b(0)+c
• 12 = a(1)^2+b(1)+c
• 16 = a(2)^2+b(2)+c
______________
• c = 10
• 12 = a+b+10
• 16 = 4a+2b+10 (then I'll divide it by two)
______________
• c = 10
• 2 = a+b
• 8-5 = 2a+b
______________
• c = 10
• a+b = 2
• 3= 2a+b
______________
• c= 10
• a = 2-b
• 3 = 2 (2-b) + b
______________
• c= 10
• a = 2-b
• 3 = 4-2b+b
______________
• c = 10
• a = 2-b
• 3-4 =-b
_______________
• c= 10
• a = 2-b
• b = 1
_______________
• c = 10
• a = 1
• b = 1

TC = Q^2+Q+10

Now, the profit is maximized when MC (Marginal Cost) equals MR,
(Marginal Revenues), which is equal to the market Price.
• If P>MC the profit increases by P-MC by producing one more unit
• If P<MC, the profit diminishes by P-MC (which is negative), but more importantly, by decreasing the quantity produced, the MC diminishes faster than the price, and the profit increases.
• If P=MC the profit could be maximum or minimum. It is maximum when the MC is increasing.

MC = 2Q+1 ( = dTC/dQ)
The total revenue function is a straight line whose equation is: TR = 7Q
MR = 7 = P

2Q+1 = 7
2Q = 6
Q = 6/2
Q = 3

NOW WE CHECK THAT THE MC IS INCREASING.

The parabola in convex, and the minimum can be found in two different ways:

-b/2a
= -1/2(1) = -1/2

OR

2Q+1>0
2Q > -1
Q > -1/2

For Q>-1/2 the parabola in increasing

If Q<0 there's no economic interpretation, so you have to consider a restriction of the domain to (0, + infinity), so the solution we found (Max profit --> Q=3) is a possible solution.

The economic profit can be found as TR-TC(Q), and it is 7*3-22 = 21-22 = -1
*NOTICE THAT* the economic profit is always negative for every level of Q.
However, if the firm didn't produce, it'd have to pay 10€ (or $) for she total sunk fixed costs, having a profit of -10. The profit of -1 is still a loss, but the firm loses less than by producing 0.

I don't know how to use the integers so I'll leave it to those who can do it.

I can't edit the post (I don't know why but I'll continue here).

You can maximize the profit by calculating the economic profit function:
π(Q) = PQ- (Q^2+Q+10)
π(Q) = 7Q-Q^2-Q-10
π(Q) = 6Q - Q^2 - 10

Marginal profit:
6 -2Q
Max profit:
6-2Q = 0
-2Q= -6
Q = 6/2
Q = 3

Draw a vertical line whose equation is Q0 = 3
The profit is the area between the vertical axis, the vertical line whose equation is written above and the profit curve.