The monthly demand equation for an electric utility company is estimated to be p = 76 - (10-5)x, where p is measured in dollars and x is measured in thousands of killowatt-hours. The utility has fixed costs of $9,000,000 per month and variable costs of $46 per 1000 kilowatt-hours of electricity generated, so the cost function is C(x) = 9• 106 + 46x. (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 $46 to $56, so its new cost function is C, (x) = 9• 106 + 56x. Should the utility pass all this increase of $10 per thousand kilowatt-hours on to the consumers? (a) Find the value of x and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit. x=0 (Type an integer or a decimal.) X =
The monthly demand equation for an electric utility company is estimated to be p = 76 - (10-5)x, where p is measured in dollars and x is measured in thousands of killowatt-hours. The utility has fixed costs of $9,000,000 per month and variable costs of $46 per 1000 kilowatt-hours of electricity generated, so the cost function is C(x) = 9• 106 + 46x. (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 $46 to $56, so its new cost function is C, (x) = 9• 106 + 56x. Should the utility pass all this increase of $10 per thousand kilowatt-hours on to the consumers? (a) Find the value of x and the corresponding price for 1000 kilowatt-hours that maximize the utility's profit. x=0 (Type an integer or a decimal.) X =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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