A fertilizer producer finds that it can sell its product at a price of p = 3000 – 0.6x dollars per unit when it produces x units of fertilizer. The total production cost (in dollars) for xunits is C(x) = 90,000 + 750x + 0.150x². 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 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).

A fertilizer producer finds that it can sell its product at a price of p = 3000 – 0.6x dollars per unit when it produces x units of fertilizer. The total production cost (in dollars) for xunits is C(x) = 90,000 + 750x + 0.150x². 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 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).

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter9: Quadratic Functions And Equations

Section9.4: Solving Quadratic Equations By Factoring

Problem 2GP

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