optimum value in function

a)C
= (1/8)t² - 21t + 3000
= (1/8) (t² - 168t + 24000)
= (1/8) (t² - 2(84)t + 84² + 24000 - 84²)
= (1/8) ( (t - 84)² + 16944)
When t = 84 , C is a minimum. He should drive 84 hours in a month.b)The minimum operation cost of his taxi
= (1/8) (16944)
= $2118


2014-10-27 23:50:22 補充：
You cann't forget to time all of them by 1/8 after you times all of them by 8.
Times (1/8)t² - 21t + 3000 by 8
= 8(1/8)t² - 8(21t) + 8(3000)
= t² - 168t + 24000
So (1/8)t² - 21t + 3000 = (1/8) (t² - 168t + 24000).

2014-10-27 23:50:29 補充：
Note that (t - 84)² is always ≥ 0 , so the min. of C = (1/8) (0 + 16944)
i.e. (t - 84)² = 0 or t = 84.

2014-10-30 02:40:01 補充：
(1/8)(t^2-168+24000)
=(1/8)(t^2-168+84^2-84^2)+24000 is wrong!

Your calculating the step should be :
(1/8)(t² - 168t + 24000)
= (1/8)(t² - 168t + 84² - 84²) + (1/8)24000
= (1/8)( (t - 84)² - 84² ) + (1/8)24000
= (1/8) (t - 84)² - (1/8)84² + (1/8)24000
= (1/8) (t - 84)² + 2118

thanks for answering my extra question.
but...i hv some problems again when i am calculating the step as follows:
(1/8)(t^2-168+24000)
=(1/8)(t^2-168+84^2-84^2)+24000
=(1/8)(t-84^2)+23118



why cant i evaluate it as (1/8)(-84^2)+24000 ?
actually need ignore the 1/8...?

2014-10-28 20:38:15 補充：
is it usual practice that i should ignore the 1/8 ,not multiply it into the bracket ....while just evaluate -84^2+24000

2014-11-08 14:06:24 補充：
thanks for ur help! I solve them out already!!