Question 1105000
The total energy in megawatt-hour used by a town is given by

E(t)=400t+(2400/π)sin(πt/12)

where t>=0 is measured in hours,with t=0 corresponding to noon.At what time of the day is the rate of energy consumption is minimum?
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{{{E(t) = 400t + (2400/pi)sin(pi*t/12)}}}
E'(t) = {{{400 + (2400/pi)cos(pi*t/12)*(pi/12)}}} = 0
Solve for t.