# Kal Tech Engineering Inc manufactures video games for "The Play Station". Variable costs are...

## Question:

Kal Tech Engineering Inc manufactures video games for "The Play Station". Variable costs are estimated to be $20 per unit and fixed costs are$10,875. The demand-price relationship for this product is Q = 1,000 - 4P where P is the unit sales price of the game and Q is the demand in number of units.

(a) Find the break-even quantity (or quantities).

(b) What is the company's maximum possible revenue?

(c) What profit would the company obtain by maximizing its total revenue?

(d) What is the company's maximum possible profit?

## Alternative Firm Strategies

A firm will typically price its products so as to maximize profits. If the firm's cost and revenue functions are known, it is possible to compare the implications of that strategy with alternative strategies such as maximizing revenues, minimizing costs or simply breaking even.

(a) The break-even quantity is 45.7 or, rounding up, 46 units.

The break-even quantity occurs where the average total cost (ATC) curve cuts the demand curve, which reflects average revenue for different quantities demanded. The break-even point is determined as follows:

• {eq}\begin{align} ATC &= AFC + AVC\\ \\ &= \frac{10,875}{Q} + 20\\ \\ Q &= 1,000 - 4P\\ \\ P &= \frac{1,000}{4} - \frac{Q}{4}\\ \\ &= 250 - \frac{Q}{4}\\ \\ ATC &= P\\ \\ \frac{10,875}{Q} + 20 &= 250 - \frac{Q}{4}\\ \\ \frac{43,500}{Q} + 80 &= 1,000 - Q\\ \\ 43,580 &= 1,000Q - Q^2\\ \\ Q &= 45.665. \end{align} {/eq}

The value for Q in the second last expression above can be derived in a variety of ways, including algebraically, or numerically using either a trial and error iterative approach or a program function such as Excel's "Solver" function.

(b) The maximum total revenue is 62,500. The company's maximum possible total revenue (TR) occurs at the point where marginal revenue (MR) becomes equal to 0. It is determined as follows: • {eq}\begin{align} P &= 250 - \frac{Q}{4}\\ \\ TR &= (P)(Q)\\ \\ &= (250 - \frac{Q}{4})(Q)\\ \\ &= 250Q - \frac{Q^2}{4}\\ \\ MR &= \frac{d TR}{d Q}\\ \\ &= 250 - \frac{2Q}{4}\\ \\ &= 250 - \frac{Q}{2}\\ \\ 0 &= 250 - \frac{Q}{2}\\ \\ \frac{Q}{2} &= 250\\ \\ Q &= 500.\\ \\ TR &= (P)(Q)\\ \\ &= (250 - \frac{Q}{4})(500)\\ \\ &= (250 - \frac{500}{4})(500)\\ \\ &= (125)(500)\\ \\ &= 62,500. \end{align} {/eq} (c) The profit the company would obtain by maximizing total revenue is41,625

When TR is maximized, profit ({eq}\pi {/eq}) would be calculated as follows, where TC is total cost:

• {eq}\begin{align} \pi &= TR -TC\\ \\ &= 62,500 - (ATC)(Q)\\ \\ &= 62,500 - (\frac{10,875}{Q} + 20)(500)\\ \\ &= 62,500 - (\frac{10,875}{500} + 20)(500)\\ \\ &= 62,500 - (41.75)(500)\\ \\ &= 62,500 - 20,875\\ \\ &= 41,625. \end{align} {/eq}

(d) The company's maximum possible profit is \$42,025.

The maximum possible profit occurs at the point where MR = MC. It is determined as follows:

• {eq}\begin{align} MR &= MC\\ \\ 250 - \frac{Q}{2} &= 20\\ \\ \frac{Q}{2} &= 230\\ \\ Q &= 460\\ \\ \pi &= TR -TC\\ \\ &= (P)(Q) - (ATC)(Q)\\ \\ &= (250 - \frac{460}{4})(460) - (\frac{10,875}{460} + 20)(460)\\ \\ &= (135)(460) - (43.6413)(460)\\ \\ &= 62,100 - 20,075\\ \\ &= 42,025. \end{align} {/eq}