A fertilizer producer finds that it can sell its product at a price of p = 1000 – 0.2x dollars per unit when it produces x units of fertilizer. The total production cost (in dollars) for x units is C(x) = 30,000 + 250x + 0.050x². If the production capacity of the firm is at most 1000 units of fertilizer in a specified time, how many units must be manufactured and sold in that time to maximize the profit? Selling i units maximizes profit.

Algebra for College Students
10th Edition
ISBN:9781285195780
Author:Jerome E. Kaufmann, Karen L. Schwitters
Publisher:Jerome E. Kaufmann, Karen L. Schwitters
Chapter8: Functions
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A fertilizer producer finds that it can sell its product at a price of p = 1000 – 0.2x dollars per unit when it produces x units of
fertilizer. The total production cost (in dollars) for x units is
C(x) = 30,000 + 250x + 0.050x².
If the production capacity of the firm is at most 1000 units of fertilizer in a specified time, how many units must be manufactured and
sold in that time to maximize the profit?
Selling i
units maximizes profit.
eTextbook and Media
Assistance Used
Hint
Assistance Used
Create a function, R(x), that models the revenue from producing x units of fertilizer. The profit can then be expressed as
P(x) = R(x) – C(x).
Transcribed Image Text:A fertilizer producer finds that it can sell its product at a price of p = 1000 – 0.2x dollars per unit when it produces x units of fertilizer. The total production cost (in dollars) for x units is C(x) = 30,000 + 250x + 0.050x². If the production capacity of the firm is at most 1000 units of fertilizer in a specified time, how many units must be manufactured and sold in that time to maximize the profit? Selling i units maximizes profit. eTextbook and Media Assistance Used Hint Assistance Used Create a function, R(x), that models the revenue from producing x units of fertilizer. The profit can then be expressed as P(x) = R(x) – C(x).
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