Question

AQ2
Data	Units [x]	Cost [C]	Rev [R]	Profit [P]
	100	260000	47500	-212500
	250	275000	109375	-165625
	500	300000	187500	-112500
	800	330000	240000	-90000
	1000	350000	250000	-100000
	1200	370000	240000	-130000
	1300	380000	227500	-152500

As your Division’s Chief economist, you perform a periodic review of the current total Cost "C", Revenue "R" and Profit "P" models associated with one of the Division's newest, but possibly underperforming, product lines. Among other important questions, you would like answers to the ones provided below. [the dependent variables are “C”, “R” or “P”; the independentvariable is “x”, the number of units (produced and) sold]. The "best fit" linear or quadratic “C”, “R” or “P” models, derived from the data referenced above, follow:

    C = 100(x)¹ + 250,000 with [(R^2)] = 1.0.

    R = –0.25(x)² + 500(x)¹ with [(R^2)] = 1.0.

    P = –0.25(x)² + 400(x)¹ – 250,000 with [(R^2)] = 1.0.  

1. Calculate how many product units “x” should be sold per sales period to maximizethis product’s total Profit “P”…then…calculate “P_max” at this “x” value. Assume market constraints suggest that 1,400 is the maximum number of product units that actually can be sold per sales period (i.e. 0 < x < 1,400 units).  

2. If internal production or external market constraints are not an issue, determine, if possible, the minimum number of product units “x” that must be produced and sold in order to “break even” (i.e. generate a Profit of $0.00)?

Answer 1

1. The profit would be maximum where

$\frac{\mathrm{d} P}{\mathrm{d} x} = 0$

or

-(-0.25 + 400r- 250000) = 0 dr %3D

or

or

or units.

Hence, profit would be maximum when production level is 800 units, which satisfies the constraint 0<x<1400.

The maximum profit would be as or (dollars). The negative profit indicates that the loss would be minimum at x=800, and the loss would be 90000.

2. The profit when maximum is negative, which implies that the profit never reaches $0. The maximum profit is -$90000, and since it is non-positive, the profit is never zero. Also, we have the the average cost as $AC = \frac{C}{x} = 100 + \frac{250000}{x}$ , which is asymptotic to the AC=100 horizontal line. Hence, theoretically, AC would be minimum where $x = \infty$ , and minimum AC would be 100 in that case.

150 AC 100 50- 200000 300000 400000 500000 100000

As the profit never reaches zero, and since AC is minimum for a very large unattainable amount of x, it is not possible to determine the breakeven amount of x.