Kal Tech Engineering Inc manufactures video games for \"The Play Station\". Vari
ID: 1217396 • Letter: K
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 $10,875. The demand-price relationship for this product is Q = 1,000 - 4P where” P” is the unit sales price of the game and “Q” is the demand in number of units. (a) Find the breakeven quantity (or quantities). (b) What is the company's maximum possible revenue? (c) What profit would the company obtain by maximizing its total revenue? (d) What is the company's maximum possible profit?
Explanation / Answer
Q = 1,000 - 4P
4P = 1,000 - Q
P = 250 - 0.25Q
(a) In break-even,
Revenue = Costs
P x Q = V x Q + F, where V: Unit variable cost & F: Fixed cost
250Q - 0.25Q2 = 20Q + 10,875
0.25Q2 - 230Q + 10,875 = 0
Solving this quadratic equation using online solver, we get
Q = 870 or Q = 50
(b) Revenue, TR = P x Q = 250Q - 0.25Q2
Revenue is maximum when dTR / dQ = 0
250 - 0.5Q = 0
0.5Q = 250
Q = 500
So, Revenue is maximized when Q = 500
(c) When Q = 500, P = 250 - (0.25 x 500) = 250 - 125 = 125
Profit ($) = Revenue - Variable cost - Fixed cost = (P x Q) - (V x Q) - F = Q x (P - V) - F
= 500 x (125 - 20) - 10,875 = 500 x 105 - 10,875 = 52,500 - 10,875 = 41,625
(d) Profit is maximized when
dTR / dQ = V
250 - 0.5Q = 20
0.5Q = 230
Q = 460
P = 250 - (0.25 x 460) = 250 - 115 = 135
Maximum Profit = Q x (P - V) - F = 460 x (135 - 20) - 10,875 = 460 x 115 - 10,875 = 52,900 - 10,875 = 42,025
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