Chapter 2, Problem 22P

To determine

(a)

The equation for Total cost and total revenue.

Explanation of Solution

Given information:

P = -0.25D + 250

P = unit sales price

D = annual quantity demanded for the product.

Total Cost (TC) = Fixed Cost+ Variable cost

Total Revenue (TR) − Quantity demanded X Price

Total Profit = Total Revenue − Total Costs

The cost and revenue details of the company are given below:

The variable costs per unit are $20.00 and the fixed costs are $10.875.00

The price-demand relationship is P = -0.25D+250 .......(1)

Here,

P = unit sales price

D = annual quantity demanded for the product.

The following three formulas were also provided to assist you with the problem as wll.

Total Cost (TC) = Fixed Cost+ Variable cost

Total Revenue (TR) − Quantity demanded× Price

Total Profit = Total Revenue − Total Costs

To determine the total cost of the problem, it should be noted that TC is equal to the combination of both fixed costs and variable costs.

Total Costs (TC) - $10,875.00+20.00D

Where, D is the quantity demanded or sold.

Total Revenue can be calculated by the multiplying price with quantity

Total Revenue = Price × Quantity Demanded

=(-0.25D+250)×D

= -0.25D² + 250D

Total Revenue = -0.25D² + 250D.

To determine

(b)

Breakeven quantity.

Explanation of Solution

Breakeven point for units is that one would need to produce inorder to achieve no profit or loss. In this case, set the formula,

Total Cost = Total Revenue

$10,875+20.00D = -0.25D² +250D

0.25D² -250D+20.00D+$10,875.00=0

0.25D² -230D+$10,875.00=0

The standard formula that assists in getting the correct answer.

$\begin{array}{l}\text{D=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2a}}\\ \text{D=}\frac{\text{-(-230)±}\sqrt{{\text{230}}^{\text{2}}\text{-4}\times \text{0}\text{.25}\times \text{10,875}}}{\text{2}\times \text{0}\text{.25}}\\ \text{=}\frac{\text{+230±}\sqrt{\text{52,900-10,875}}}{\text{0}\text{.5}}\\ \frac{\text{230±}\sqrt{\text{42,025}}}{\text{0}\text{.5}}\\ \frac{\text{230±205}}{\text{0}\text{.5}}\\ \text{D=870}\text{or}\text{50}\end{array}$

Thus, the break even points for this situation are both 50 units and 870 units respectively. It can be said that anything above 870 units and below 50 units could constitute a loss whereas anything between would be profitable to profitable.

To determine

(c)

Number of units the company would want to produce if they wanted to maximize the revenues.

Explanation of Solution

Total Revenue = -0.25D² +250D

$\begin{array}{l}\text{MR=}\frac{\text{dTR}}{\text{dD}}\text{=-2×0}{\text{.25D}}^{\text{2-1}}\\ \text{MR=-0}\text{.5D+250}\end{array}$

Revenue maximization:

Revenue is maximized when MR equals to zero.

$MR=-0.5D+250$

Revenue maximization;

$\begin{array}{l}\text{MR=-0}\text{.5D+250=0}\\ \text{0}\text{.5D}\text{=}\text{250}\\ \text{D}\text{=}\frac{\text{250}}{\text{0}\text{.5}}\\ \text{=500}\text{units}\end{array}$

Thus to maximize revenue, the company must produce 500 units.

To determine

(d)

Total Profit at revenue maximization point of 500 units:

Explanation of Solution

Total Profit = Total Revenue- Total Costs

$\begin{array}{l}=(-\$0.25D^2+\$250D)-(\$10,875+20\times 500)\\ (-\$62,500+\$125,000)-(\$20,875)\\ =\$62,500-\$20,875\\ =\$41,625\end{array}$

Thus, the total profit at revenue maximization point of 500 units is $41,625.

The company’s maximum profit is obtained when marginal revenue equals to marginal cost.

$\begin{array}{l}\text{Total Revenue=-\$0}{\text{.25D}}^{\text{2}}\text{+\$250D}\\ \text{MR}\text{=}\frac{\text{dTR}}{\text{dD}}\text{=-2×0}{\text{.25D}}^{\text{2-1}}{\text{+250D}}^{\text{1-1}}\\ \text{MR}\text{=}\text{-0}\text{.5D+250}\\ \end{array}$

Marginal Cost (MC) is calculated as shown below:

$\begin{array}{l}\text{Total}\text{Costs}\text{(TC)=\$10,875}\text{.00+20}\text{.00D}\\ \text{MC}\text{=}\frac{\text{dTC}}{\text{dD}}\text{=0+20}{\text{.00D}}^{\text{1-1}}\\ \text{MC}\text{=}\text{20}\end{array}$

Profit maximization is at MR = MC

$\begin{array}{l}\text{-0}\text{.5D+}\text{250=20}\\ \text{0}\text{.5D}\text{=}\text{250-20}\\ \text{=230}\\ \text{=}\frac{\text{230}}{\text{0}\text{.5}}\\ \text{D}\text{=}\text{460}\text{units}\end{array}$

Thus, maximum profit is obtained at 460 units.

Total Profit = Total Revenue − Total Costs

$\begin{array}{l}\text{=(-\$0}{\text{.25D}}^{\text{2}}\text{+\$250D)-(\$10,875+20D)}\\ \text{Given}\text{that}\text{D=460}\text{units}\\ \text{=(-\$0}{\text{.25(460)}}^{\text{2}}\text{+\$250×460)-(\$10,875+20×460)}\\ \text{=(-\$52,900+\$115,000)-(\$20,075)}\\ \text{=\$62,100-\$20,075}\\ \text{=\$42,025}\\ \end{array}$

Maximum possible profit is $42,025.

To determine

(e)

Graphical Representation:

Explanation of Solution

Graphical Representation: