In May 2006. Commonwealth Edison Company suplied electricityto residences for a

Question

In May 2006. Commonwealth Edison Company suplied electricityto residences for a monthly customer charge of $7.58 plus 8.275cents per kilowatt-hour (kWhr) for the first 400kWhr supplied inthe month and 6.208 cents per kWhr for all usage over 400 kWhr forall usage over 400kWhr in the month. (a) what is the charge for using 300 kWhr in a month? (b) what is the charge for using 700 kWhr in a month? (c) If C is the monthly charge for x kWhr. develop a modelrelating the monthly charge and kilowatt-hours used. That is,express C as a function of x. In May 2006. Commonwealth Edison Company suplied electricityto residences for a monthly customer charge of $7.58 plus 8.275cents per kilowatt-hour (kWhr) for the first 400kWhr supplied inthe month and 6.208 cents per kWhr for all usage over 400 kWhr forall usage over 400kWhr in the month. (a) what is the charge for using 300 kWhr in a month? (b) what is the charge for using 700 kWhr in a month? (c) If C is the monthly charge for x kWhr. develop a modelrelating the monthly charge and kilowatt-hours used. That is,express C as a function of x.

Explanation / Answer

(a) If someone used 300kWhr, then they are chargeda rate of 8.275 cents for every 1kWhr: 300(8.275)= 2482.5 cents but that's in cents, so: 2482.5/(100)= $24.825 round $24.83 plus the $7.58 for being a customer $24.83+$7.58= $32.41 (b) 700dWhr is more than 400 so the first 400 equals: 400(8.275) times 100 to make it dollars 400(8.275)/(100) Plus the remaining 300kWhr 400(8.275)/(100) + 300(6.208)/(100) Plus the charge of being a customer 400(8.275)/(100) + 300(6.208)/(100) + 7.58 =59.304 =$59.30 (c)By evaluating the steps taken in a and be we can derive 2functions and create a piecewise function. f(x)= 7.58 + (8.275)x/100          if x400 , x>0 This can be simplified to: f(x)= 7.58 + (8.275)x/100           if x400 , x>0 Simplified again to: f(x)= 7.58 + (8.275)x/100               if x400 , x>0

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