## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 1 - Section 1.7 - Combinations of Functions; Composite Functions - Exercise Set - Page 259: 99

#### Answer

$(R-C)(20000)=-200,000$ dollars; loses $200,000$ dollars when selling 20,000 radios. $(R-C)(30000)=0$ dollars; breaks even when selling 30000 radios. $(R-C)(40000)=200,000$ dollars; makes a profit of $200,000$ dollars when selling 40,000 radios.

#### Work Step by Step

Step 1. Given the equations for the cost and revenue, we have the annual profit as $P(x)=R(x)-C(x)=65x-(600000+45x)=20x-600000=20(x-30000)$ dollars Step 2. For $x=20000$, we have $P(20000)=(R-C)(20000)=20(20000-30000)=-200,000$ dollars, which means that the company loses $200,000$ dollars when selling 20,000 radios. Step 3. For $x=30000$, we have $P(30000)=(R-C)(30000)=20(30000-30000)=0$ dollars, which means that the company breaks even when selling 30000 radios. Step 4. For $x=40000$, we have $P(40000)=(R-C)(40000)=20(40000-30000)=200,000$ dollars, which means that the company makes a profit of $200,000$ dollars when selling 40,000 radios.

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