At most, Alana can spend [tex]$\$[/tex] 40[tex]$ on carnival tickets. Ride tickets cost $[/tex]\[tex]$ 4$[/tex] each, and food tickets cost [tex]$\$[/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, $[/tex]f$, she buys.

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

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

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



Answer :

To solve the problem and find the maximum number of ride tickets [tex]\( r \)[/tex] Alana can buy, let's follow a structured approach:

1. Define the variables:
- Let [tex]\( r \)[/tex] represent the number of ride tickets.
- Let [tex]\( f \)[/tex] represent the number of food tickets.

2. Set up the inequalities based on the problem description:
- Alana needs to buy at least a total of 16 tickets:
[tex]\[ r + f \geq 16 \][/tex]
- Alana has at most \[tex]$40 to spend where ride tickets cost \$[/tex]4 each and food tickets cost \$2 each:
[tex]\[ 4r + 2f \leq 40 \][/tex]

3. Simplify the inequalities:
- Simplify the second inequality by dividing through by 2:
[tex]\[ 2r + f \leq 20 \][/tex]

4. Consider the boundary conditions:
- To find the feasible region, let's convert the inequalities to equations:
[tex]\[ r + f = 16 \][/tex]
[tex]\[ 2r + f = 20 \][/tex]

5. Solve the system of equations:
- Set the equations equal to each other by substituting [tex]\( f \)[/tex] from the second equation into the first:
[tex]\[ f = 20 - 2r \][/tex]
Substitute [tex]\( f \)[/tex] into the first equation:
[tex]\[ r + (20 - 2r) = 16 \][/tex]
Simplify and solve for [tex]\( r \)[/tex]:
[tex]\[ r + 20 - 2r = 16 \][/tex]
[tex]\[ -r + 20 = 16 \][/tex]
[tex]\[ r = 4 \][/tex]

6. Verify the solution:
- Substitute [tex]\( r = 4 \)[/tex] back into one of the original inequalities to ensure it satisfies both conditions:
[tex]\[ r + f = 4 + f \geq 16 \quad \text{implies} \quad 4 + f = 16 \quad \text{so}\quad f = 12 \][/tex]
[tex]\[ 2r + f = 2 \times 4 + 12 = 8 + 12 = 20 \quad \text{(which is also less than 20)} \][/tex]

Thus, the maximum number of ride tickets Alana can buy is indeed [tex]\( 4 \)[/tex].

So, the answer is [tex]\(\boxed{4}\)[/tex].