Last time we developed data showing that our greatest profits come when we produce between 40 and 50 units; that is, when 40 <= Q <= 50. Now we want to answer the question: How can we find the exact value of Q that gives the greatest p? One way is to just keep zooming in on the area of the data. In the next table, we zoom in on the area between 40 and 50 to see what happens:
Q p R C p
50 5.0 250.0 150 100.0
49 5.1 249.9 149 100.9
48 5.2 249.6 148 101.6
47 5.3 249.1 147 102.1
46 5.4 248.4 146 102.4
45 5.5 247.5 145 102.5
44 5.6 246.4 144 102.4
43 5.7 245.1 143 102.1
42 5.8 243.6 142 101.6
41 5.9 241.9 141 100.9
40 6.0 240.0 140 100.0
The area of greatest profit appears to be at Q = 45. But suppose it's at Q = 45.1? Or Q = 44.9? We could zoom in again to find out. In fact we can repeatedly zoom in as far as we want, since there's no shortage of real numbers. Another way to find the Q for greatest profit is to incrementally change Q. We can do this with the concept of marginal profit.
Marginal profit (Mp) is computed as: Mp = (p2 - p1)/(Q2 - Q1). That is, for a change in Q, how much does p change? From the data, we see that the change in p varies with the change in Q. We do not have a constant rate of change. This is because the profit equation yields a curve, not a straight line, as we saw when we developed our model previously. We would need to compute Mp for every point we want to check.
Here is an example for a change in Q from 50 to 49 from the data above:
Mp = (p2 - p1)/(Q2 - Q1)
Mp = (100 - 100.9)/(50 - 49)
Mp = -.9/1
Mp = -.9
Here is another example for a change in Q from 46 to 45:
Mp = (p2 - p1)/(Q2 - Q1)
Mp = (102.4 - 102.5)/(46 - 45)
Mp = -.1/1
Mp = -.1
So, given that the profit equation does not give us a straight line, how can we identify the Q which yields the greatest p? We can see a clue by noticing that there is a definite maximum in profit curve, where it reached the top of its' arc before it begins to descend.
This is the point where an additional change in Q results in Mp = 0, and further changes cause Mp to become negative. So the Q which gives the greatest p lies at the top of the curve where Mp = 0. If we were to draw a tangent line at that point, its' slope would be zero:
So if we can find the point on the curve where the slope of a tangent line is zero, we will have found the point of greatest profit and consequently the desired Q. How do we do that? Differential calculus.
Recall that the profit equation is: p = -.1Q^2 + 9Q - 100. It turns out that by taking the derivative of this equation with respect to Q, we can find the slope of the tangent line at any point along the curve. Here's how:
Step 1: Find the derivative of the profit equation. This gives us the marginal profit:
p = -.1Q^2 + 9Q - 100
Mp = dp/dQ = -.2Q + 9
Now for any Q, we can find the marginal profit.
Step 2: Since we are looking for the point where Mp = 0, and we now know Mp = dp/dQ = -.2Q + 9, we can set -.2Q + 9 = 0 and solve for Q:
Mp = dp/dQ = -.2Q + 9 = 0
-.2Q = -9
Q = -9/-.2
Q = 45
So our profit maximizing Q is 45. What price is required to sell this quantity?
Recall that p = 10 - .1Q, so p = 10 - .1(45) = 5.5
What is the profit we can expect to earn for Q = 45 and p = 5.5?
Recalling that R = pQ, C = 100 + Q, and p = R - C, we see that:
R = pQ = 45 * 5.5 = 248
Using the revenue equation we developed previously, we can confirm:
R = -.1Q^2 + 10Q = 248
C = 100 + Q = 100 + 45 = 145
p = R - C = 248 - 145 = 103
p = -.1Q^2 + 9Q - 100 = -.1(45^2) + 9(45) - 100 = 103
Now that we have R and C, we can also find our break-even point
. That's the subject of the next section.