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
中四數 - 部分變~! 急~!!!!!!!!!!!!!!!!!!!!!!
2007-01-08 3:53 am
回答 (2)
2007-01-08 4:22 am
✔ 最佳答案
(a )P = kx + hx^2, where h,k =/= 0When 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
2007-01-08 4:29 am
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
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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