Solving the Revenue Word Problem
Now that we can set up these problems, we are ready to solve them. For each of the previous examples and problems, a desired revenue will be given. We will set that revenue equal to the revenue equation. This will be a quadratic equation. Some of these equations will be solved by factoring, others by the quadratic formula. Some problems will have more than one solution.
Examples
Example 1:
A department store sells 20 portable stereos per week at $80 each. The manager believes that for each decrease of $5 in the price, six more stereos will be sold.
Let x represent the number of $5 decreases in the price.
P = 80 – 5 x Q = 20 + 6 x R = (80 – 5 x )(20 + 6 x ).
What price should be charged if the revenue needs to be $2240?
R = (80 – 5 x )(20 + 6 x ) becomes 2240 = (80 – 5 x )(20 + 6 x )
If x = , the price for each stereo will be .
If x = 2, the price for each stereo will be P = 80 – 5(2) = $70.
Example 2:
A rental company manages an office complex with 16 offices. Each office can be rented if the monthly rent is $1000. For each $200 increase in the rent, one tenant will be lost.
Let x represent the number of $200 increases in the rent.
P = 1000 + 200 x Q = –16 – 1 x R = (1000 + 200 x )(16 – x )
What should the monthly rent be if the rental company needs $20,800 each month in revenue?
If x = 3, the monthly rent will be 1000 + 200(3) = $1600. If x = 8, the monthly rent will be 1000 + 200(8) = $2600.
Example 3:
A grocery store sells 300 pounds of bananas each day when they are priced at 45 cents per pound. The produce manager observes that for each 5cent decrease in the price per pound of bananas, an additional 50 pounds are sold.
Let x represent the number of 5cent decreases in the price.
P = 45 – 5 x Q = 300 + 50 x R = (45 – 5 x )(300 + 5 x )
What should the price of bananas be for weekly sales to be $140? How many bananas (in pounds) will be sold at this price (these prices)?
(The revenue will be in terms of cents, so $140 becomes 14,000 cents.)
If x = 2, the price per pound will be 45 – 5(2) = 35 cents. The number of pounds sold each week will be 300 + 50(2) = 400. If x = 1, the price per pound will be 45 – 5(1) = 40 cents and the number of pounds sold each week will be 300 + 50(1) = 350.
Example 4:
A music storeowner sells 60 newly released CDs per week when the cost is $12 per CD. For each $1.50 decrease in the price, the store will sell an additional 16 CDs per week.
Let x represent the number of $1.50 decreases in the price.
P = 12.00 – 1.50 x Q = 60 + 16 x R = (12.00 – 1.50 x )(60 + 16 x )
What should the price be if the storeowner needs revenue of $810 per week for the sale of these CDs? How many will be sold at this price (these prices)?
When x = 1.25, the price should be 12 – 1.50(1.25) = $10.13 and the number sold would be 60 + 16(1.25) = 80. If x = 3, the price should be 12 – 1.50(3) = $7.50 and the number sold would be 60 + 16(3) = 108.
Find practice problems and solutions at Revenue Practice Problems  Set 2.
More practice problems for this concept can be found at: Algebra Quadratic Applications Practice Test.
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