Solve using the quadratic formula

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SUMMARY

The discussion focuses on maximizing revenue for a car dealership by adjusting the markup price using the quadratic formula. The dealership currently sells cars at a $6000 markup, averaging 80 sales per month. With each $100 increase in markup, one less car is sold. The revenue function is established as (80-x)(6000+100x), which simplifies to -100x² + 2000x + 480000. The maximum revenue of $490,000 occurs when x equals 10, indicating a new markup of $6100.

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  • Understanding of quadratic equations and their properties
  • Familiarity with revenue functions in economics
  • Knowledge of the quadratic formula for solving equations
  • Basic concepts of markup pricing and sales volume relationships
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A large dealership had been selling new cars at $6000 over the factory price. Sales have been averaging 80 cars per mouth. Because of inflation, the $6000 markup is going to be increased. The marketing manager has determined that, for every $100 increase, there will be one less car
sold each month. What should the new markup be in order to maximize the revenue?

Let x rep the number of $100 increases
the equation for revenu is (#sold)(price)
so (80-x)(6000+100x)
then, when you expand it its -100x^2+2000x+480 000
next, solve using the quadratic formula
is everything right?

Thanks
 
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Why would you find the values of x giving zero profit?

Find the MAXIMUM of the given profit function, not its roots.
 
^ok, thanks, so if my work abouve was correct, the answer should be a maximum of 490 000, when x=10
 
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