(a) D^{*}=\frac{a-c_{v}}{2 b}=\frac{\$ 180-\$ 83}{2(0.02)}=2,425 units per month [from Equation (2-10)].
\left(a-c_{v}\right)>0 ?
($180 − $83) = $97, which is greater than 0.
And is (total revenue − total cost) > 0 for D^{*} = 2,425 units per month?
[$180(2,425) − 0.02(2,425)2] − [$73,000 + $83(2,425)] = $44,612
A demand of D^{*} = 2,425 units per month results in a maximum profit of $44,612 per month. Notice that the second derivative is negative (−0.04).
(b) Total revenue = total cost (breakeven point)
-b D^{2}+\left(a-c_{v}\right) D-C_{F} =0 \quad \text { [from Equation (2-11)] } \\-0.02 D^{2}+(\$ 180-\$ 83) D-\$ 73,000 =0 \\-0.02 D^{2}+97 D-73,000 =0
* Given the quadratic equation ax^{2}+bx+c=0, the roots are given by x=\frac{-b\mp \sqrt[]{b^{2}-4ac} }{2a}
And, from Equation (2-12),
(2-12)
D^{\prime}=\frac{-\left(a-c_{v}\right) \pm\left[\left(a-c_{v}\right)^{2}-4(-b)\left(-C_{F}\right)\right]^{1 / 2}}{2(-b)}
D^{\prime}=\frac{-97 \pm\left[(97)^{2}-4(-0.02)(-73,000)\right]^{0.5}}{2(-0.02)} \\D_{1}^{\prime}=\frac{-97+59.74}{-0.04}=932 \text { units per month } \\D_{2}^{\prime}=\frac{-97-59.74}{-0.04}=3,918 \text { units per month. }
Thus, the range of profitable demand is 932–3,918 units per month
Spreadsheet Solution
Figure 2-5(a) displays the spreadsheet solution for this problem. This spreadsheet calculates profit for a range of demand values (shown in column A). For a specific value of demand, price per unit is calculated in column B by using Equation (2-1) and Total Revenue is simply demand × price. Total Expense is computed by using Equations (2-7) and (2-8). Finally, Profit (column E) is then Total Revenue − Total Expense
(2-7)
C_{T}=C_{F}+C_{V,}
(2-8)
C_{V}=c_{v}\cdot D,
A quick inspection of the Profit column gives us an idea of the optimal demand value as well as the breakeven points. Note that profit steadily increases as demand increases to 2,500 units per month and then begins to drop off. This tells us that the optimal demand value lies in the range of 2,250 to 2,750 units per month. A more specific value can be obtained by changing the Demand Start point value in cell E1 and the Demand Increment value in cell E2. For example, if the value of cell E1 is set to 2,250 and the increment in cell E2 is set to 10, the optimal demand value is shown to be between 2,420 and 2,430 units per month.
The breakeven points lie within the ranges 750–1,000 units per month and 3,750–4,000 units per month, as indicated by the change in sign of profit. Again, by changing the values in cells E1 and E2, we can obtain more exact values of the breakeven points.
Figure 2-5(b) is a graphical display of the Total Revenue, Total Expense, and Profit functions for the range of demand values given in column A of Figure 2-5(a). This graph enables us to see how profit changes as demand increases. The optimal demand value (maximum point of the profit curve) appears to be around 2,500 units per month.
Figure 2-5(b) is also a graphical representation of the breakeven points. By graphing the total revenue and total cost curves separately, we can easily identify the breakeven points (the intersection of these two functions). From the graph, the range of profitable demand is approximately 1,000 to 4,000 units per month. Notice also that, at these demand values, the profit curve crosses the x-axis ($0).
Comment
As seen in the hand solution to this problem, Equations (2-10) and (2-12) can be used directly to solve for the optimal demand value and breakeven points.
The power of the spreadsheet in this example is the ease with which graphical displays can be generated to support your analysis. Remember, a picture really can be worth a thousand words. Spreadsheets also facilitate sensitivity analysis (to be discussed more fully in Chapter 11). For example, what is the impact on the optimal demand value and breakeven points if variable costs are reduced by 10% per unit? (The new optimal demand value is increased to 2,632 units per month, and the range of profitable demand is widened to 822 to 4,443 units per month.)
