Question

The cost, in dollars, of operating a machine per day is given by the formula C=2t^2-84t+1025, where t is the time, in hours, the machine operates. What is the minimum cost of running the machine? For how many hours must the machine run to reach this minimum cost?

Answer 1

The minimum time is where t=1 because 1 hour is the least time, hence substitute this value into the polynomial. \[C=2*1-84*1+1025=943\] I'm pretty sure that's the answer.

Answer 2

If they mean working for a full day, then you will have to substitute 24 hours (a day) into the equation; hence: \[C=2(24*24)-(84*24)+1025=1152-2016+1025=161\] that is probably the minimum because it is less than the original one.

Answer 3

@Sepeario at the back of my textbook it says the minimum cost of $143 occurs when the machine runs for 21 hours

Answer 4

so...i really don't know what to write

Answer 5

\[c=2t^2-84t+1025\]
\[\frac{ dc }{ dt }=4t-84\]
\[\frac{ dc }{ dt }=0\]
gives
4t-84=0
t=21
\[\frac{ d^2c }{ dt^2 }=4\]
at t=21
\[\frac{ d^2c }{ dt^2 }=4>0\]
c is minimum at t=21
so
\[minimum~ c=2(21)^2-84*21+1025=?\]

Answer 6

Or you can also solve by completing the squares.