Derivative Questions

Question

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

 

E(t)=400t+(2400/pi)sin(pi.t/12)

 

where t>=0 is measured in hours,with t=0 corresponding o noon.At what time of the day is the rate of energy consumption is minumum?

Answer 1

If we reformulate the question, find a value of t for which function E(t) reaches min.

 

Take derivative of E(t)

 

E'(t)=400+(2400/pi)cos(pi.t/12)*pi/12=

=400+200cos(pi.t/12)

 

Now E'(t)=0 ⇒ 400+200cos(pi.t/12)=0 

Solve the last equation for t and and figure out that value of t for which E(t) reaches min. I think, the rest obvious.

 

400+200cos(pi.t/12)=0

cos(pi.t/12)=-1/2

pi.t/12=ArcCos(-1/2)+2pik, k  from Integers

pi.t/12=2pi/3+2pik, 

t=12*2/3+12*2k

t=8+24k, k=...-3.-2,-1,0, +1, +2, +3,..

 

Now it is easy to see that k take values 0,1,2,,, (why?)

for

   k=0, t=8

 k=1, t=32...