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

(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 is $41,625

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}