Business Calculus

AQ1
X410	Units [x]	Cost [C]	Rev [R]	Profit [P]
	100	218750	42500	-176250
	250	257188	96875	-160313
	500	308750	162500	-146250
	800	350000	200000	-150000
	1000	365000	200000	-165000
	1200	370000	180000	-190000
	1300	368750	162500	-206250

One day, the Division Director asks you to informally review the current total Cost “C”, Revenue “R” and Profit “P” models associated with associated with Product line “###”, the Division’s newest, but possibly underperforming, product line.  Among other important questions, the Director would like answers to ones provided below.  [The dependent variables are “C”, “R” or “P”; the independent variable is “x”, the number of units (produced and) sold].  Quadratic “C”, “R” or “P” models provide the “best fit” to the data on which the these models (referenced above), are based.  These models follow:

    C = –0.125(x)2 + 300(x)1 + 190,000 with [(R^2)] = 1.0.

    R = –0.25(x)2 + 450(x)1 with [(R^2)] = 1.0.

    P = –0.125(x)2 + 150(x)1 – 190,000 with [(R^2)] = 1.0.

a. Calculate how many product units “x” should be sold per sales period to maximize this product’s total Profit “P”…then…calculate “Pmax” 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 per sales period.

Solution

For 2nd order, optimum value exists is on vertex.

The product units that will maximize profit , x= -b/2a, where b=150 and a = -0.125.

In other words, the vertex of equation ,

1) The optimum profit will be at x = 600

So,

2) 600 units lies in the range , below 1400 units , and therefore , the optimum profit will be at x = 600

 

b. If internal production or external market constraints were 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). 

Solution;

At break –even point, total revenue equals to total costs and thus P(x) = 0.

Solving for x , using quadratic formula

The value under the square root is -72500, and the roots will be imaginary numbers.

Therefore , it is not possible to find break even points.