Max values of a function
\[E(t) = \frac{ 13500 }{ t+1 }\] \[L(t)=\frac{ 4500t+250 }{ t+1}\]
\[126000 \] gallons at t=0
E(t) and L(t) represent rates at which water enters/leaves the lake.
Please help!
can you take derivative ?
I didnt think i had to bc the functions represent rates.
to get max or min value, you need to equate the first derivative of the functions to 0
Right but the functions already represent the derivative bc they're rates. Maybe im mistaken. But thats what ive been taught.
ohh...can you post the entire question ?
Sure, gimme a sec to type please.
During a recent storm, water entered a lake through a stream and left through a spillway over a dam. The water entered at a rate given by the function \[E(t)=\frac{ 13500 }{ t+1 }\] and left the lake at a rate given by the function \[L(t)=\frac{ 4500t+250 }{ t+1 }\].. Both rates are in terms of gallons per hour and \[t\] is time in hours. At \[t=0\]the lake contained \[126000\] gallons of water.
I did parts a,b, and c.
Part d: At what time does the lake reach it maximum volume? Part e: What is the maximum volume of the lake?
so, to get max. volume, e(t) should be maximum and l(t) should be minimum, right ??
max, water should enter lake and min water shouldd leave the lake, to get max volume, make sense ? and yes, since they are rates, you'll just put t=0 in them.
Ohh, that simple?
So E(0) = 13500, L(0) =250
no, actually, sorry. rates are to be equated to 0 so, e(t) =0 and l(t) = 0 actually, increase in volume will be proportional to e(t) - l (t) so, equate that to 0
do you have answers to verify ?
Thats what I did before. I did E(t)-L(t)=0, thus E(t)=L(t), when t= 2.944
isn't that correct ?
I wasnt sure.
well, i get the same. so you can be sure now.
Thanks. Brb
part e) max volume = 126000+e(2.944)-l(2.944) see whether this makes sense.
Ohh, hmm..ill try that.
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