a couple are celebrating their 20th anniversary by having a reception at a local reception hall. they have…
a couple are celebrating their 20th anniversary by having a reception at a local reception hall. they have budgeted $2,000 for their reception. if the reception hall charges a $90 cleanup fee plus $34 per person, find the greatest number of people that they may invite and still stay within their budget. the couple can invite at most people to the reception (round down to the nearest whole person)
Answer
Explanation:
Step1: Set up cost - equation
Let $x$ be the number of people. The total cost $C$ is given by the equation $C = 90+34x$. The couple has a budget of $2000$. So we set up the inequality $90 + 34x\leq2000$.
Step2: Solve the inequality for $x$
First, subtract 90 from both sides of the inequality: $34x\leq2000 - 90$. So $34x\leq1910$. Then, divide both sides by 34: $x\leq\frac{1910}{34}\approx56.176$.
Step3: Round down to the nearest whole person
Since we need to stay within the budget and $x$ represents the number of people, we round down the result. So $x = 56$.
Answer:
56