The cost in dollard, of operating a machine per day is given in the formula C=3t^2 -96t +1024 , where t is the time the machine operates, in hours. What is the minimum cost of running the machine? for how many hours must the machine run to reach the minimum cost?
yes but how do I use it thats what Im confused about
use derivatives to find the turning point of the equation
I need this question for today badly and I cannot figure it out for my life
\(\large C=3t^2−96t+1024\) \(\large C'=6t−96\) when C' = 0... its a turning point \(\large C'=6t−96\) \(\large 0=6t−96\) \(\large 96=6t\) \(\large t=96/6\)
so turning point is at t = 16?
ok thanks but how many hours does it need?
so sub t = 16 back into \(\large C=3t^2 -96 +1024\) to find the minimum cost (at t = 16hrs)
and that would give me the minimum cost?
yep
YOU ARE A GEM
so howd u go? minimum cost is...?
sec
1696
...? nuh uh...?
so is that right?
3(16^2)-96 +1024
no, sorry i got a completely different answer, are u sure u did minus not plus for the 2nd term?
3(16^2) - 96(16) + 1024
ohh i forgot the t on 96
256
sweeeet, purfect!
AYYYY
can you help me with another one?
aye, i'll try
A triangle has base 2x+ 1 and height 6x-3. What value of x would give an area of 240m^2?
k, so tell me formula for area of triangle?
A=hbb 2
also, ur triangle look like which pic...?|dw:1402925538599:dw|
the thing is its a word question no pics
or there is another question
k, so assume one on left |dw:1402925681214:dw|
\(\Large A~ =\frac {~ base~ \times ~height}2\)
can u solve from here...?
(1/2)(2x+1)(6x-3) = 240 multiplying both sides by 2: (2x+1)(6x-3) = 480 FOIL the left side: 12x^2 - 6x + 6x - 3 = 480 12x^2 - 3 = 480 12x^2 = 483 x^2 = 483/12 x = +/- (6.3443) but for measurements we only have positive therefore x=6.3443
is this right?
ywah, looks perfect, gimme a sec to check tho
yep perfect, i kept leaving the 1/2 part out, my bad ;)
last one?
please
shoot
the sum of squares of two consecutive integers in 365 . find the value of the intergers.
x^2 + (x+1)^2 = 365... expand out the brackets and solve for x
my teacher teach crap very well. can you show me?
the damm keyboard. * my teacher dosen't tach very well. Could you show me
teach*
x^2 + (x+1)^2 = 365 x^2 + (x+1)(x+1) = 365 x^2 + (x^2+1x+ 1x+1) = 365 x^2 + x^2 + 2x + 1 = 365 2x^2 + 2x + 1 = 365 2x^2 + 2x + 1 - 365 = 0 now use quadratic equation:
so in this case: a = 2 b = 2 c = -364
so if u plug those values into the quadratic formula, you'll get x = -14 or x = 13... so x = 13 sounds much more realistic so ur 2 consecuative values are x and x+1 so 13 and 14
hold on. Isnt c 1?
nope, 1-365 = -364 you always have to arrange the equation so it = 0, or it doesnt work properly hey, it's pretty much midnight in oz, catchya and good luck gaz ;)
imma sleep now, sorry
OMG THNX
np, slaters
GIVE YOU A NOBLE PRIZE FOR THAT
lol, blushes ;) nah, all good hey, good luck with ur study
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