The monthly demand equation for an electric utility company is estimated to be p equals 59 minus (10 Superscript negative 5 Baseline )x​, wh

Question

The monthly demand equation for an electric utility company is estimated to be p equals 59 minus (10 Superscript negative 5 Baseline )x​, where p is measured in dollars and x is measured in thousands of​ killowatt-hours. The utility has fixed costs of ​$3 comma 000 comma 000 per month and variable costs of ​$27 per 1000​ kilowatt-hours of electricity​ generated, so the cost function is Upper C (x )equals 3 times 10 Superscript 6 Baseline plus 27 x. ​(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​ utility’s variable costs from ​$27 to ​$41​, so its new cost function is Upper C 1 (x )equals 3 times 10 Superscript 6 Baseline plus 41 x. Should the utility pass all this increase of ​$14 per thousand​ kilowatt-hours on to the​ consumers?

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Nevaeh 1 week 2021-10-08T01:24:00+00:00 1 Answer 0

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    2021-10-08T01:25:16+00:00

    Answer:

    a) The price that maximizes profit is p =$43.

    b) The utility shouldn’t pass all this increase on the consumers because this would mean a decrease in the profits. They should pass only $7 to consumers price.

    Step-by-step explanation:

    We have an electric utility company which has a demand function defined by:

    p=59-10^{-5}x

    where p is the price and x is the the energy in thousands of kWh.

    The cost of the company is defined as:

    C(x)=3\cdot 10^6+27x

    We have to calculate the price that maximizes the utility’s profit R(x).

    We can define the profit as:

    R(x)=p\cdot x-C=(59-10^{-5}x)\cdot x-(3\cdot 10^6+27x)\\\\R(x)=-10^{-5}x^2+(59-27)x-3\cdot10^6\\\\R(x)=-10^{-5}x^2+32x-3\cdot10^6.

    To maximize R, we have to derive it and equal to zero

    \dfrac{dR}{dx}=0\\\\\\\dfrac{dR}{dx}=-2\cdot10^{-5}x+32=0\\\\\\x=(32/2)\cdot 10^5=16\cdot 10^5=1.6\cdot 10^6

    The price that maximizes the profit is then:

    p=59-10^{-5}x=59-10^{-5}(16\cdot 10^5)=59-16=43

    b) When the unit cost rise from $27 to $41, the utility profit function changes.

    We have to calculate the new price that maximizes profit, and then we will know if the rise in cost was transferred to price completely.

    The new profit function is:

    R(x)=p\cdot x-C=(59-10^{-5}x)\cdot x-(3\cdot 10^6+41x)\\\\R(x)=-10^{-5}x^2+(59-41)x-3\cdot10^6\\\\R(x)=-10^{-5}x^2+18x-3\cdot10^6

    To maximize R, we have to derive it and equal to zero

    \dfrac{dR}{dx}=0\\\\\\\dfrac{dR}{dx}=-2\cdot10^{-5}x+18=0\\\\\\x=(18/2)\cdot 10^5=9\cdot 10^5

    We calculate the new price as:

    p=59-10^{-5}x=59-10^{-5}(9\cdot 10^5)=59-9=50

    The new price is $7 dollars above the previous maximizing price, so the rise in unit cost is only transferred 50% to the price.

    The utility shouldn’t pass all this increase on the consumers because this would mean a decrease in the profits. They should pass only $7 to consumers price.

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