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Revenue Help

By — McGraw-Hill Professional
Updated on Sep 27, 2011

Introduction to Revenue Word Problems

A common business application of quadratic equations occurs when raising a price results in lower sales or lowering a price results in higher sales. The obvious question is what to charge to bring in the most revenue. This problem is addressed in Algebra II and Calculus. The problem addressed here is finding a price that would bring in a particular revenue.

The problem involves raising (or lowering) a price by a certain number of increments and sales decreasing (or increasing) by a certain amount for each incremental change in the price. For instance, suppose for each increase of $10 in the price, two customers are lost. The price and sales level both depend on the number of $10 increases. If the price is increased by $10, two customers are lost. If the price is increased by $20, 2(2) = 4 customers will be lost. If the price is increased by $30, 2(3) = 6 customers will be lost. If the price does not change, 2(0) = 0 customers will be lost. The variable will represent the number of incremental increases (or decreases) of the price.

Setting up the Revenue Formula

The revenue formula is R = PQ where R represents the revenue, P represents the price, and Q represents the number sold. If the price is increased, then P will equal the current price plus the variable times the increment. If the price is decreased, then P will equal the current price minus the variable times the increment. If the sales level is decreased, then Q will equal the current sales level minus the variable times the incremental loss. If the sales level is increased, then Q will equal the current sales level plus the variable times the incremental gain.

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. Then the price will decrease by 5 x :

P = 80 – 5 x .

The sales level will increase by six for each $5 decrease in the price—the sales level will increase by 6 x :

Q = 20 + 6 x .

R = PQ becomes R = (80 – 5 x )(20 + 6 x ).

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 )

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 5-cent decrease in the price per pound of bananas, an additional 50 pounds are sold.

Let x represent the number of 5-cent decreases in the rent.

P = 45 – 5 x Q = 300 + 50 x R = (45 – 5 x )(300 + 5 x )

(The revenue will be in cents instead of dollars.)

Example 4:

A music storeowner sells 60 newly released CDs per day when the price is $12 per CD. For each $1.50 decrease in the price, the store will sell an additional 16 CDs each 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 )

Find practice problems and solutions at Revenue Practice Problems - Set 1.

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