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Lionel uses two functions to model the costs of having a problem diagnosed and repaired by two appliance repair companies.
He models the cost, in dollars, of having Phirs Appliances handle the repair job, requiring $x$ hours of labor, with function [tex]p : p(x)=68 x+81[/tex].
He models the cost of having Quality Electric handle the same repair with function [tex]q : q(x)=51 x+90[/tex].
The function represents the amount, in dollars, Lionel will save by having Quality Electric handle an [tex]x[/tex]-hour repair instead of Phir's Appliances.
Lionel will save If he uses Quality Electric for a repair needing 2 hours of labor.
Answer (2)
Define the savings function as the difference between the costs of the two companies: s ( x ) = p ( x ) − q ( x ) .
Substitute the given cost functions: s ( x ) = ( 68 x + 81 ) − ( 51 x + 90 ) .
Simplify the savings function: s ( x ) = 17 x − 9 .
Calculate the savings for a 2-hour repair: s ( 2 ) = 17 ( 2 ) − 9 = 25 . The savings is 25 .
Explanation
Problem Analysis Let's analyze the problem. We have two appliance repair companies, Phirs Appliances and Quality Electric, with their costs modeled by the functions p ( x ) = 68 x + 81 and q ( x ) = 51 x + 90 , respectively, where x is the number of hours of labor. We want to find the function that represents the savings Lionel will get by choosing Quality Electric over Phirs Appliances, and then calculate the savings for a 2-hour repair job.
Finding the Savings Function First, we need to find the savings function, which is the difference between the cost of Phirs Appliances and the cost of Quality Electric. Let's call this savings function s ( x ) . So, we have s ( x ) = p ( x ) − q ( x )
Substituting the Cost Functions Now, substitute the given functions p ( x ) and q ( x ) into the savings function: s ( x ) = ( 68 x + 81 ) − ( 51 x + 90 )
Simplifying the Savings Function Simplify the savings function: s ( x ) = 68 x + 81 − 51 x − 90 s ( x ) = ( 68 x − 51 x ) + ( 81 − 90 ) s ( x ) = 17 x − 9 So, the savings function is s ( x ) = 17 x − 9 .
Calculating Savings for 2-Hour Repair Next, we need to calculate the savings for a 2-hour repair. To do this, we substitute x = 2 into the savings function: s ( 2 ) = 17 ( 2 ) − 9
Evaluating the Savings Evaluate s ( 2 ) :
s ( 2 ) = 34 − 9 s ( 2 ) = 25 So, Lionel will save $25 if he uses Quality Electric for a 2-hour repair.
Examples
Imagine you're deciding between two phone repair services. One charges a higher hourly rate but a lower initial fee, while the other has a lower hourly rate but a higher initial fee. By creating functions to model each service's cost, you can determine which is more cost-effective for different repair durations. This is similar to Lionel's situation, where he's comparing appliance repair services. Understanding these cost differences helps in making informed decisions, whether it's for phone repairs, appliance services, or any situation where you're weighing different pricing structures.
Lionel will save $25 by using Quality Electric for a 2-hour repair instead of Phirs Appliances. This is calculated by determining the savings function as s ( x ) = 17 x − 9 and substituting x = 2 . Therefore, s ( 2 ) equals $25.
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