Question

The monthly demand equation for an electric utility company is estimated to be p=49-(10⁻⁵) x, where p is measured in dollars and x is measured in thousands of killowatt-hours. The utility has fixed costs of ​$1,000,000 per month and variable costs of ​$40 per 1000​ kilowatt-hours of electricity​ generated, so the cost function is

C(x)=1•10⁶+40x.
Find the value of ( x ) and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit.

Answer 1

To find the value of (x) and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit, calculate the profit function by subtracting the cost function from the revenue function. Differentiate the profit function and set it equal to 0 to solve for x. Plug the value of x back into the demand equation to find the corresponding price.

Step-by-step explanation:

To find the value of (x) and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit, we need to calculate the profit function.

Profit, P(x), is equal to revenue minus cost. Revenue is given by the equation R(x) = px, where p is the price per kilowatt-hour. Substitute the demand equation p=49-(10⁻⁵)x into the revenue equation to get R(x) = (49-(10⁻⁵)x) x. The cost function is C(x)=1x10⁶+40x.

Profit is then calculated by subtracting the cost function from the revenue function: P(x) = R(x) - C(x).

To maximize profit, differentiate the profit function with respect to x and set it equal to 0. Differentiate R(x) - C(x) and set it equal to 0 to solve for x. Plug the value of x back into the demand equation to find the corresponding price.