Chapter 11 QuaDraTiC appliCaTionS 391
- Let x = first number and x + 7 = second number.
xx
xx
xx
xx
()
()()
+=
+=
+−=
−+=
7 228
7 228
7 228 0
12 19 0
2
2
x
x
x
−=
=
+=
12 0
12
719
x
x
x
+=
=−
+=−
19 0
19
712
There are two solutions: 12 and 19, −12 and −19.
Revenue 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 obvi-
ous question is what price brings in the most revenue. This problem is addressed
in pre-calculus and calculus. The problem addressed here is finding a price that
would bring some specific revenue.
These revenue problems involve 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(1) = $10, two
customers are lost. If the price is increased by $2(10) = $20, 2(2) = 4 customers
will be lost, and if the price is increased by $3(10) = $30, 2(3) = 6 customers
will be lost. In general, if the price increases by $10x, then 2x customers will be
lost. The variable represents the number of incremental increases (or decreases)
of the price.
For the following problems, we use the revenue formula, R = PQ, where R
represents the revenue, P represents the price, Q represents the number sold, and
x represents the number of incremental increases or decreases in the price. If the
price is increased, then P is the current price plus the x times the increment. If
the price is decreased, then P is the current price minus x times the increment.
If sales decrease, then Q is the current sales level minus x times the incremental
loss. If sales increase, then Q is the current sales level plus x times the incremental
change. For now, we focus on how to represent the revenue in terms of x.
