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在一間書店裡，某詞典每本的售價是＄x，將該詞典全部賣出的利潤是＄P，已知P的其中一部分隨 x 正變，而另一部分則隨 x ^ 2 正變，當 x = 35 時, P = 78750 ; 又當 x = 60 時, P = 60000 a, 試以 x 表示 P b, 若該詞典每本的售價是 $ 65 , 求利潤 c, 求將該詞典全部賣出所能賺取的最大利潤, 以及所對應該詞典每本的售價

最佳解答:

(a )P = kx + hx^2, where h,k =/= 0 When x = 35, P = 78750 (78750) = k(35) + h(35)^2 2250 = k + 35h......(1) When x = 60, P = 60000 (60000) = k(60) + h(60)^2 1000 = k + 60h......(2) (1) - (2): 1250 = -25h h = -50 Sub. h = -50 into (2) 1000 = k + 60(-50) k = 4000 ∴ P = 4000x - 50x^2 (b) Sub. x = 65 into P = 4000x - 50x^2 P = 4000(65) - 50(65)^2 = 48750 ∴利潤是＄48750 (c) P = 4000x - 50x^2 = -50(x^2 - 80x) = -50(x^2 - 80x + 40^2 - 40^2) = -50(x^2 - 80x + 40^2) + 80000 = -50(x - 40)^2 + 80000 ∴最大利潤是＄80000 所對應該詞典每本的售價是＄40

其他解答:

a) P = ax + bx^2 x = 35, P = 78750 78750 = 35a + 1225b ... (1) x = 60, P = 60000 60000 = 60a + 3600b ... (2) 12(1) - 7(2): 525000 = -10500b b = -50 a = 4000 P = 4000x - 50x^2 b) x = 65 P = 4000(65) - 50(65)^2 P = 48750 利潤是$48750 c) A.maths method 學左微分 P = 4000x - 50x^2 P' = 4000 - 100x set P' = 0 0 = 4000 - 100x x = 40 P = 4000(40) - 50(40)^2 P = 80000 maths method P = 4000x - 50x^2 max P = (4ac - b^2) / 4a = -(4000)^2 / 4(-50) = 80000 80000 = 4000x - 50x^2 50x^2 - 4000x + 80000 = 0 x^2 - 80x + 1600 = 0 (x - 40)^2 = 0 x = 40
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