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標題:
Easy Math
發問:
A store has been selling a product at the price of $40 per unit, and at this price, players have been buying 50 units per month. The owner of the store wishes to raise the price of the game and estimates that for each dollar increase in price, three fewer units will be sold each month. If each unit costs the store... 顯示更多 A store has been selling a product at the price of $40 per unit, and at this price, players have been buying 50 units per month. The owner of the store wishes to raise the price of the game and estimates that for each dollar increase in price, three fewer units will be sold each month. If each unit costs the store $25, at what price should the game be sold to maximize profit? Ans:$41
最佳解答:
Let the increasing price is x. Then the profitP = (40 + x)(50 - 3x) - 25(50 - 3x) = (15 + x)(50 - 3x)P' = -3(15 + x) + (50 - 3x)Set P' = 050 - 3x = 45 + 3xx = 5/6However, as x should be integer, we need to consider x = 0 and x = 1when x = 0, profit = 15 * 50 = 750when x = 1, profit = 16 * 47 = 752So, the game should be sold at $41 to maximize profit
$25 => 95units => (25 - 25) x 95 = $0 $26 => 92units => (26 - 25) x 92 = $92 $27 => 89units => (27 - 25) x 89 = $178 $28 => 86units => (28 - 25) x 86 = $258 $29 => 83units => (29 - 25) x 83 = $332 $30 => 80units => (30 - 25) x 80 = $400 $31 => 77units => (31 - 25) x 77 = $462 $32 => 74units => (32 - 25) x 74 = $518 $33 => 71units => (33 - 25) x 71 = $568 $34 => 68units => (34 - 25) x 68 = $612 $35 => 65units => (35 - 25) x 65 = $650 $36 => 62units => (36 - 25) x 62 = $682 $37 => 59units => (37 - 25) x 59 = $708 $38 => 56units => (38 - 25) x 56 = $728 $39 => 53units => (39 - 25) x 53 = $742 $40 => 50units => (40 - 25) x 50 = $750 $41 => 47units => (41 - 25) x 47 = $752 $42 => 44units => (42 - 25) x 44 = $748 $43 => 41units => (43 - 25) x 41 = $738 $44 => 38units => (44 - 25) x 38 = $722 $45 => 35units => (45 - 25) x 35 = $700 $46 => 32units => (46 - 25) x 32 = $672 $47 => 29units => (47 - 25) x 29 = $638 $48 => 26units => (48 - 25) x 26 = $598 $49 => 23units => (49 - 25) x 23 = $552 $50 => 20units => (50 - 25) x 20 = $500 $51 => 17units => (51 - 25) x 17 = $442 $52 => 14units => (52 - 25) x 14 = $378 $53 => 11units => (53 - 25) x 11 = $308 $54 => 8units => (54 - 25) x 8 = $232 $55 => 5units => (55 - 25) x 5 = $150 $56 => 2units => (56 - 25) x 2 = $62 thus, the price should be sold at $41 per unit to maximize profit.
Easy Math
發問:
A store has been selling a product at the price of $40 per unit, and at this price, players have been buying 50 units per month. The owner of the store wishes to raise the price of the game and estimates that for each dollar increase in price, three fewer units will be sold each month. If each unit costs the store... 顯示更多 A store has been selling a product at the price of $40 per unit, and at this price, players have been buying 50 units per month. The owner of the store wishes to raise the price of the game and estimates that for each dollar increase in price, three fewer units will be sold each month. If each unit costs the store $25, at what price should the game be sold to maximize profit? Ans:$41
最佳解答:
Let the increasing price is x. Then the profitP = (40 + x)(50 - 3x) - 25(50 - 3x) = (15 + x)(50 - 3x)P' = -3(15 + x) + (50 - 3x)Set P' = 050 - 3x = 45 + 3xx = 5/6However, as x should be integer, we need to consider x = 0 and x = 1when x = 0, profit = 15 * 50 = 750when x = 1, profit = 16 * 47 = 752So, the game should be sold at $41 to maximize profit
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其他解答:$25 => 95units => (25 - 25) x 95 = $0 $26 => 92units => (26 - 25) x 92 = $92 $27 => 89units => (27 - 25) x 89 = $178 $28 => 86units => (28 - 25) x 86 = $258 $29 => 83units => (29 - 25) x 83 = $332 $30 => 80units => (30 - 25) x 80 = $400 $31 => 77units => (31 - 25) x 77 = $462 $32 => 74units => (32 - 25) x 74 = $518 $33 => 71units => (33 - 25) x 71 = $568 $34 => 68units => (34 - 25) x 68 = $612 $35 => 65units => (35 - 25) x 65 = $650 $36 => 62units => (36 - 25) x 62 = $682 $37 => 59units => (37 - 25) x 59 = $708 $38 => 56units => (38 - 25) x 56 = $728 $39 => 53units => (39 - 25) x 53 = $742 $40 => 50units => (40 - 25) x 50 = $750 $41 => 47units => (41 - 25) x 47 = $752 $42 => 44units => (42 - 25) x 44 = $748 $43 => 41units => (43 - 25) x 41 = $738 $44 => 38units => (44 - 25) x 38 = $722 $45 => 35units => (45 - 25) x 35 = $700 $46 => 32units => (46 - 25) x 32 = $672 $47 => 29units => (47 - 25) x 29 = $638 $48 => 26units => (48 - 25) x 26 = $598 $49 => 23units => (49 - 25) x 23 = $552 $50 => 20units => (50 - 25) x 20 = $500 $51 => 17units => (51 - 25) x 17 = $442 $52 => 14units => (52 - 25) x 14 = $378 $53 => 11units => (53 - 25) x 11 = $308 $54 => 8units => (54 - 25) x 8 = $232 $55 => 5units => (55 - 25) x 5 = $150 $56 => 2units => (56 - 25) x 2 = $62 thus, the price should be sold at $41 per unit to maximize profit.
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