Question
The library is holding a used book sale, and the head librarian wants to know how much to sell each book for in order to make $150. Based on a previous sale, the librarian can use the expression
–
10p+80 to estimate the number of books that will be sold based on the sales price, p.
What two sales prices will bring the library exactly $150 in revenue?
$
or $
–
10p+80 to estimate the number of books that will be sold based on the sales price, p.
What two sales prices will bring the library exactly $150 in revenue?
$
or $
Answers
GPT-4o mini
To find the two sales prices that will bring in exactly $150 in revenue, we have the expression for the number of books sold, which is given by:
\[
n(p) = -10p + 80
\]
Where \( n(p) \) is the number of books sold based on the sales price \( p \).
The revenue \( R \) generated from selling the books can be calculated using the formula:
\[
R = \text{number of books sold} \times \text{price per book}
\]
Substituting \( n(p) \) into the revenue formula, we get:
\[
R = (-10p + 80)p
\]
We want to find when the revenue \( R \) equals $150:
\[
(-10p + 80)p = 150
\]
Expanding the left side:
\[
-10p^2 + 80p = 150
\]
Rearranging this into a standard quadratic equation:
\[
-10p^2 + 80p - 150 = 0
\]
To simplify, divide the entire equation by -10:
\[
p^2 - 8p + 15 = 0
\]
Now we can factor this quadratic equation. We look for two numbers that multiply to +15 and add up to -8. These numbers are -3 and -5, hence we can factor the quadratic as:
\[
(p - 3)(p - 5) = 0
\]
Setting each factor equal to zero gives us the possible sales prices:
\[
p - 3 = 0 \quad \Rightarrow \quad p = 3
\]
\[
p - 5 = 0 \quad \Rightarrow \quad p = 5
\]
Thus, the two sales prices that will bring the library exactly $150 in revenue are:
\[
\boxed{3} \text{ and } \boxed{5}
\]
\[
n(p) = -10p + 80
\]
Where \( n(p) \) is the number of books sold based on the sales price \( p \).
The revenue \( R \) generated from selling the books can be calculated using the formula:
\[
R = \text{number of books sold} \times \text{price per book}
\]
Substituting \( n(p) \) into the revenue formula, we get:
\[
R = (-10p + 80)p
\]
We want to find when the revenue \( R \) equals $150:
\[
(-10p + 80)p = 150
\]
Expanding the left side:
\[
-10p^2 + 80p = 150
\]
Rearranging this into a standard quadratic equation:
\[
-10p^2 + 80p - 150 = 0
\]
To simplify, divide the entire equation by -10:
\[
p^2 - 8p + 15 = 0
\]
Now we can factor this quadratic equation. We look for two numbers that multiply to +15 and add up to -8. These numbers are -3 and -5, hence we can factor the quadratic as:
\[
(p - 3)(p - 5) = 0
\]
Setting each factor equal to zero gives us the possible sales prices:
\[
p - 3 = 0 \quad \Rightarrow \quad p = 3
\]
\[
p - 5 = 0 \quad \Rightarrow \quad p = 5
\]
Thus, the two sales prices that will bring the library exactly $150 in revenue are:
\[
\boxed{3} \text{ and } \boxed{5}
\]