1. We know that for each $5 decrease in price, the school can sell 100 more yearbooks. This means that if x represents the number of $5 decreases in price, then the total number of yearbooks sold will be 500 + 100x.
2. The price of each yearbook will also decrease by $5 for each decrease in price, so the new price will be $50 - $5x.
3. The revenue from yearbook sales is the product of the number of yearbooks sold and the price per yearbook. Therefore, the revenue function can be represented as R(x) = (500 + 100x)(50 - 5x).
4. Expanding the expression gives R(x) = 500(50) + 500(-5x) + 100x(50) + 100x(-5x) = 25000 - 2500x + 5000x - 500x^2
5. Simplifying further, we get R(x) = -500x^2 + 2500x + 25000. Comparing this expression to R(x) = (500+ax)(50-bx), we can see that a = 100 and b = 5.
Therefore, a = 100 and b = 5.
A school typically sells 500 yearbooks each year for $50 each. The economics class does a project and discovers that they can sell 100 more yearbooks for every $5 decrease in price. The revenue for yearbook sales is equal to the number of yearbook sold times the price of the yearbook. Let x represent the number of $5 decreases in price. If the expression that represents the revenue is in the form R(x)=(500+ax)(50-bx). Find the values of a and b.
a=100
b=5
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