求救maths~~~~~~更正之前錯誤題~~

Question

Question :
The total cost to produce x units of a low-tech product is given by C (X) =120+6x
and the total revenue is given by R (X) = 48x-3x^2----. Find
a) the break-even quantity
b) the maximum possible profit ( profit means revenue - cost )
c) the quanity for a profit of $3000

我唔識　ｂ　呀~~48x-3x^2 - 120+6x= profit啦
但點搵個maximum profit 呀?? 個x 係幾多呀?? 點搵個x呀??~~~thx~~~~~

可以有解釋嗎？？
THX A LOT FOR YOUR HELP~^^~

ｓｏｒｒｙ呀～～出現左ｄ符號．．還是不明白～～
各位高人出手幫忙幫忙～～ｔｈｘ　ａｒ～～

Answer 1

a.
Break even is when profit=0;
thus
P(x)=R(x)-C(x)=0
Two solutions are possible, namely x=4 or x=10

b.
Maximum or minimum is when P'(x)=0
P'(x)=dP(x)/dx=42-6x
The maximum profit is when P'(x)=0, or x=7, and the profit is P(7)=27
To verify that it is indeed a maximum, check P''(7)<0
P''(x)=-6x
P''(7)=-42<0, therefore it is a maximum.

c.
In general, the quantity required is obtained by solving for x in
P(x)=3000.
However, in b., we established that the maximum profit is 27, so a real value of x does not exist for a profit of 3000.

Answer 2

a)
Break even occurs when revenue = cost;
R(x) = C(x)
48x - 3x^2 = 120+6x
3x^2 - 42x + 120 = 0
x^2 - 14x +40=0
(x-10)(x-4) = 0
Therefore, x=4 or x=10

b)
Let Profit P(x) = R(x)-C(x)
P(x) = -3x^2 +42x -120
P(x) = -3(x^2-14x+49) +3(49)-120
P(x) = -3(x-7)^2 +27
The maximum profit occurs when x = 7 and the maximum profit is 27.

c)
Since the maximum profit is 27, it is impossible to get a profit of 3000 no matter in what quantity.

2006-11-09 11:34:00 補充：
For part (b), since -3(x-7)^2 is always negative unless x = 7 which results in 0Therefore, maximum of -3(x-7)^2 is 0 and the maximum of -3(x-7)^2 + 27 is 27