Night Timers is a small company manufacturing glow-in-the-dark products. One of the hottest items the engineering department has developed is adhesive tape that can be applied to walls and floors. Night Timers' chief engineer anticipates that the product will be sold in ten-foot rolls. At present, the company's maximum production capacity is 140,000 rolls per year. The engineer believes the cost function to be described by: C = $50,000 +0.25 Q

Night Timers' president seeks to establish a price that maximizes profit. She thinks that the firm should be able to sell at least 125,000 rolls of tape per year. The marketing manager forecasts demand for the tape to be: Q = 350,000 - 200,000P.

Calculate the following, assuming that Night Timers maximizes profit:

a. Quantity

b. Price

c. Total cost

d. Total revenue

This problem is very similar to the War Games problem, except this problem has a capacity limit for production of 140,000 units per year, which may come into play, and the terms are stated in C and Q instead of P and C. It is easy to switch terms. For example, in this one, the quantity and price relationship Q = 350,000 – 200,000 P can be stated in terms of P instead of Q by simple algebra:

Q = 350,000 – 200,000 P

Add 200,000 P to both sides

Q + 200,000 P = 350,000

Subtract Q from both sides 200,000

P = 350,000 – Q

Divide both sides by 200,000

P = 350,000/200,000 – Q/200,000

P = 1.75 – 0.000005 Q

Now the terms are in similar form to the first problem. I do this to demonstrate that this is the same as Problem 1, (with different values), and I need Price stated this way. Also, it helps to understand the relationships between the terms.

To solve this problem let’s review what we know. We want to maximize profit. We know that

Profit = Revenue – Costs which we can restate as

Profit = (P*Q) – ($50,000 + $0.25 Q)

This tells us that fixed costs are $50,000 and variable costs are $0.25 per unit. We need to spread the $50,000 fixed costs over as many units as we can until the marginal revenue falls below the $0.25 marginal cost. The price starts at $1.75 at Q = 0, and price drops a penny for every 2,000 units produced, or $0.000005 for each unit produced.

This is a linear relationship. The revenue from the first unit produced is $1.749995. If we sell 2, the price drops for both to $1.749990, and total revenue is $3.499980 which is an increase of $1.749985 (which is $0.00001 less than the first one). If we sell 3, revenue increases another $1.749975. In this problem, the marginal revenue is a little less for each additional unit sold, by the same increment at every level – it drops by $0.00001 for each unit sold. This is a linear function all the way out. So – how many units do we have to sell to drive marginal revenue down to the $0.25 marginal cost level? We can calculate that:

Q where MR = MC is ($1.749995 - $0.25)/($0.00001) = $1.49995/$0.00001 = 149,995 units. Hey – that is more than production capacity! Therefore, the best we can do is full production of 140,000 units. Now we need to calculate the price, revenue, costs and profits associated with full production.

At 140,000 units, with

P = 1.75 – 0.000005 Q

P = 1.75 – (0.000005 * 140,000)

P = 1.75 – 0.70 P = $1.05 $1.05

Revenue is quantity times price or

140,000 * $1.05 = $147,000

Costs are $50,000 + $0.25 * 140,000

C = $50,000 + $35,000

C = $85,000

Profit = $147,000 – $85,000 = $62,000