Question 6.22: The demand function for a good is given by the equation P = ......

(a)     (i)  TR = PQ = (50 − 2Q)Q = 50Q − 2Q²

(ii)  π = TR − TC = 50Q − 2Q² − (160 + 2Q)
= −2Q² + 48Q − 160
= −2(Q² − 24Q + 80)

(b) (i) Sketching the total cost function: TC = 160 + 2Q is a linear function, therefore two points are required to sketch its graph. For example, at Q = 0, TC = 160 + 2(0) = 160 and taking any other Q-value, say Q = 20, then TC = 160 + 2(20) = 200. The total cost function is drawn in Figure 6.24.

Sketching the total revenue function: This total revenue function is the same as that sketched in Worked Example 6.21, Figure 6.23. TR and TC are graphed in Figure 6.24.
(ii) From the graph, break-even occurs when TC = TR, at Q = 4 and Q = 20.
The break-even points are calculated algebraically as follows:

TC = TR
160 + 2Q = 50Q − 2Q²
2Q² − 48Q + 160 = 0
Q² − 24Q + 80 = 0    dividing both sides by 2
(Q − 4)(Q − 20) = 0

Therefore the break-even quantities are Q = 4 and Q = 20.
As shown in Figure 6.24, losses are incurred below break-even point Q = 4 (where TC > TR) and above break-even point Q = 20 (where TC > TR). Profits are made between the two break-even points since TC < TR in this interval.

(c) Maximum profits are calculated by finding the turning point(s) for the profit function. This is done using the three-step max/min method:

Step 1: Get derivatives

π = −2Q² + 48Q − 160

\frac {dπ}{dQ} = −4Q + 48

\frac {d^2π}{dQ^2}=-4      at turning point

Step 2: Find turning point(s)

\frac {dπ}{dQ} =0    at turning point

−4Q + 48 = 0

Q = \frac {48}{4} =12

Step 2a: At Q = 12,

π = −2(12)² + 48(12) −160 = 128

Step 3: Max or min? The second derivative is a negative constant (−4) therefore profits are a maximum π = 128 at Q = 12

The profit function has only one turning point and this is a maximum. Sketching the profit function: The profit function is sketched in Figure 6.25. To sketch the profit function, find

(1) the points of intersection with the axes and (2) the turning point.
(1) The graph cuts the Q-axis when π = 0.

π = 0
−2(Q² − 24Q + 80) = 0
Q² − 24Q + 80 = 0    dividing both sides by −2
(Q − 4)(Q − 20) = 0
therefore        Q = 4    and Q = 20

These values of Q where π = 0 are also the break-even points.
(2) We have just shown that profit has a maximum value of 128 when Q = 12. This provides another point for sketching the profit function.

(d) Maximum profit of 128 occurs when Q = 12.