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The monthly demand equation for an electric utility company is estimated to be p

ID: 3183903 • Letter: T

Question

The monthly demand equation for an electric utility company is estimated to be p 70 (105)x, where p is measured in dollars and x is measured in thousands of killowatt-hours. The utility has fixed costs of $3,000,000 per month and variable costs of $34 per 1000 kilowatt hours of electricity generated, so the cost function is Cix)-3-10+34x. (a) Find the value of x and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit. b) Suppose that the rising fuel costs increase the utlity's variable costs from $34 to $40, so its new cost function is C1(x)-3-108+40x. Should the ubility pass all this increase of $6 per thousand kilowatt-hours on to the consumers? (a) Find the value of x and the corresponding price for 1000 kilowatthours that maximize the utility's profit. (Type an integer or a decimal.) The price for 1000 kilowatt hours that maximizes profit is S (b) Suppose that the rising fuel costs increase the utility's variable costs from $34 to $40, Should the utility pass all this increase of 56 per thousand kilowatt-hours on to the consumers? O No O Yes

Explanation / Answer

(a) Let us consider the revenue function is,

R(x)=x.p(x) , where p(x) is the price function

Therefore, R(x) = x(70-10-5)x = 70x-10-5x2

the profit function can be written as :

P(x)=R(x)-C(x), where C(x) is cost function

=> P(x) = 70x-10-5x2 - 3.106-34x

= -10-5x2 +36x - 3.106

Differentiating

P'(x) = -2.10-5x+36

P'(x)=0 =>  -2.10-5x+36=0

=>x=18.105

therefore, the value of x that maximizes the profit is , 18.105 thousand kilowatt-hour

Correseponding price is,

p(18.105)=70-10-5(18.105) = 52

The price of maximum profit is $52 per 1000 killowatt hour.

(b) the new profit function will be :

P1(x)=R(x)-C1(x) =70x-10-5x2 - 3.106-40x

= -10-5x2 +30x - 3.106

Differentiating

P'(x) = -2.10-5x+30

P'(x)=0 =>  -2.10-5x+30=0

=>x=15.105

therefore, the value of x that maximizes the profit is , 15.105 thousand kilowatt-hour

Correseponding price is,

p(15.105)=70-10-5(15.105) = 55

The price of maximum profit is $55 per 1000 killowatt hour.

We can see that there is a change of $3 from the previous value.Therefore, it is not advisable to pass all this increase of $6

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