At most, Alana can spend [tex]\$ 40[/tex] on carnival tickets. Ride tickets cost [tex]\$ 4[/tex] each, and food tickets cost [tex]\$ 2[/tex] each. Alana buys at least 16 tickets. The system of inequalities represents the number of ride tickets, [tex]r[/tex], and the number of food tickets, [tex]f[/tex], she buys.

[tex]\[
\begin{array}{r}
r+f \geq 16 \\
4r + 2f \leq 40
\end{array}
\][/tex]

What is the maximum number of ride tickets she can buy?

A. 4
B. 6
C. 10
D. 12



Answer :

To determine the maximum number of ride tickets [tex]\( r \)[/tex] Alana can buy while satisfying the given constraints, we need to consider the system of inequalities:

1. [tex]\( r + f \geq 16 \)[/tex] (Alana buys at least 16 tickets in total)
2. [tex]\( 4r + 2f \leq 40 \)[/tex] (Alana spends at most \$40 on tickets)

Let's go through the steps to solve these inequalities:

1. Analyze and Simplify the Budget Inequality:

We can divide the second inequality by 2 to simplify it:
[tex]\[ 4r + 2f \leq 40 \implies 2r + f \leq 20 \][/tex]

2. Identify Constraints:

We now have two inequalities:
[tex]\[ r + f \geq 16 \][/tex]
[tex]\[ 2r + f \leq 20 \][/tex]

3. Testing Options:

Let's test each option to ensure it satisfies both inequalities.

- For [tex]\( r = 4 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 4 + f \geq 16 \implies f \geq 12 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 4 + f \leq 20 \implies 8 + f \leq 20 \implies f \leq 12 \][/tex]
Both [tex]\( f \geq 12 \)[/tex] and [tex]\( f \leq 12 \)[/tex] are satisfied, so [tex]\( f = 12 \)[/tex]. Thus, [tex]\( r = 4 \)[/tex] is a valid option.

- For [tex]\( r = 6 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 6 + f \geq 16 \implies f \geq 10 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 6 + f \leq 20 \implies 12 + f \leq 20 \implies f \leq 8 \][/tex]
Here, [tex]\( f \geq 10 \)[/tex] and [tex]\( f \leq 8 \)[/tex] are conflicting, so [tex]\( r = 6 \)[/tex] is not a valid option.

- For [tex]\( r = 10 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 10 + f \geq 16 \implies f \geq 6 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 10 + f \leq 20 \implies 20 + f \leq 20 \implies f \leq 0 \][/tex]
Here, [tex]\( f \geq 6 \)[/tex] and [tex]\( f \leq 0 \)[/tex] are conflicting, so [tex]\( r = 10 \)[/tex] is not a valid option.

- For [tex]\( r = 12 \)[/tex]:
[tex]\[ r + f \geq 16 \implies 12 + f \geq 16 \implies f \geq 4 \][/tex]
[tex]\[ 2r + f \leq 20 \implies 2 \cdot 12 + f \leq 20 \implies 24 + f \leq 20 \implies f \leq -4 \][/tex]
Here, [tex]\( f \geq 4 \)[/tex] and [tex]\( f \leq -4 \)[/tex] are conflicting, so [tex]\( r = 12 \)[/tex] is not a valid option.

Therefore, after testing all given options, the maximum number of ride tickets Alana can buy while satisfying the given inequalities is [tex]\( \boxed{4} \)[/tex].