function d數　　２

Let a(q) be the profit of the company
Then a(q)=qp(q)-c(q)
=q(-4q+960)-200q-7500
=-4q^2 +760q-7500
a) The practical value of q
咁乜野叫做practical value呢,我諗間間公司都係諗住要賺錢架啦,蝕錢就唔會去做,所以應該係有利潤
即係話profit要大過0
咁即係a(q)大過0
即係-4q^2 +760q-7500大過0
q^2 -190q+1875細過0 (兩邊除負數就大過變細過, 相反亦然)

[190-{190^2 -4(1875)}]/2 細過 q 細過 [190+{190^2 -4(1875)}] {}係平方根
[190-{36100-7500}]/2 細過 q 細過 [190+{36100-7500}]/2
10.44 細過 q 細過 179.56

咁doll就一定係integer啦

所以 11 細過 q 細過 179 就應該係practical value啦

呢題其實唔係十分肯定,但應該係岩既

b) Let r(q) be the revenue
the revenue = r(q) = qp(q)
= q(-4q+960)
= -4q^2 +960q
求佢既最大值,咁就要d一d佢
dr(q)/dq = -8q+960
當 dr(q)/dq = -8q+960 = 0 個時就係最大or最細
q=120
Check埋佢係最大定最細,就要d多次
d^2 r(q)/dq = -8 即係最大 (曲面向下)

咁所以maximum revenue就會得到when q=120
咁放返落去r(q)度就即係 r(120)=-4(120)^2+960(120)
=57600
So, maximum revenue = 57600

c)求maximum profit 就即係好似b part既方法, 不過就d a(q)
da(q)/dq = -8q+760
當da(q)/dq = -8q+760 =0 個時就係最大or最細
咁q=95
Check埋佢係最大定最細,就要d多次
d^2 r(q)/dq = -8 即係最大 (曲面向下)
咁所以 95 is the level of production in which maximum profit can be obtained.

d)咁做到上面既答案,呢一part就唔會有太大既難度啦
求price就即係求p(q)
咁當q=95時, p(95)=-4(95)+960=580
So, the corresponding price is 580 for the levels of production in (c).
到profit就緊係求a(q)啦
咁當q=95時, a(95)=-4(95)^2 +760(95)-7500
=28600
So, the corresponding profit is 28600 for the levels of production in (c).

(A) the practical range of q.
q>=0
c(q)=200q+7500 >= 0 (OK!)
p(q)=-4q+960 (no need to set the restriction >=0, negative means you give others as free gift)
(b) the maximum revenue.
revenue
= price * q
= (-4q + 960)*q
= -4q^2 + 960q
= 57600 - 57600 + 960q - 4q^2..........(completing square method!)
= 57600 - (240 - 2q)^2
The max. revenue is attained when 240-2q = 0. So, q=120.
The max. revenue is 57600.
(c) the level of production in which maximum profit can be obtained.
the level of production = 120.
(d) the corresponding price and profit for the levels of production in (c).
The corresponding price
=-4*120+960=480
The profit
= revenue-cost
=57600 - [200(120)+7500]
=26100