Question

The cost function for a certain company is $C=60 x+300$ and the revenue is given by $R=100 x-0.5 x^{2} .$ Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of $x$ (production level) that will create a profit of $\$ 300$ .

Name: Problem 55, Chapter 2: College Algebra
Uploaded: 2021-01-22T01:58:18-08:00
Duration: 4 min 37 s
Channel: Oluwadamilola Ameobi
Description: The cost function for a certain company is C=60 x+300 and the revenue is given by R=100 x-0.5 x^2 . Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of 300 .

   The cost function for a certain company is $C=60 x+300$ and the revenue is given by $R=100 x-0.5 x^{2} .$ Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of $x$ (production level) that will create a profit of $\$ 300$ .

College Algebra

Jay Abramson 1st Edition

Chapter 2, Problem 55

Instant Answer

Solved by Expert Oluwadamilola Ameobi

01/22/2021

Step 1

The given cost function is $C=60x+300$ and the revenue function is $R=100x-0.5x^{2}$. So, the profit function $P$ can be written as: \[P = R - C = (100x - 0.5x^{2}) - (60x + 300)\] Show more…

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The cost function for a certain company is $C=60 x+300$ and the revenue is given by $R=100 x-0.5 x^{2} .$ Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of $x$ (production level) that will create a profit of $\$ 300$ .

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Key Concepts

Profit Function

The profit function represents the difference between revenue and cost. It is a key concept in business mathematics and is fundamental to understanding how earnings relate to production levels or sales. In many cases, especially when revenue or cost functions are quadratic, the resulting profit function is also quadratic.

Quadratic Equations

A quadratic equation is a second degree polynomial equation that typically takes the form ax² + bx + c = 0. This concept is central to algebra since it describes parabolic relationships, and there are several methods to solve them, including factoring, completing the square, and the quadratic formula.

Translating Word Problems into Equations

This concept involves formulating algebraic equations from real-world scenarios. By understanding how to express relationships and conditions (like profit being revenue minus cost) in mathematical terms, one can set up equations that model and solve practical problems.

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