Question

Difficulty: MediumTranslating and Solving Algebraic Word Problems

A food truck selling gourmet grilled cheese sandwiches has a daily fixed operating cost of 120120. Each sandwich costs 2.502.50 to make and is sold for 6.506.50. What is the minimum number of sandwiches the food truck must sell in one day to make a net profit of at least 180180?

Answer: 75 sandwiches

Answer

The food truck must sell a minimum of 75 sandwiches to make a net profit of at least $180.
Representing the number of sandwiches sold as xx, the total revenue is 6.50x6.50x and the total cost is 120+2.50x120 + 2.50x. The net profit is the difference between revenue and cost: 6.50x(120+2.50x)6.50x - (120 + 2.50x), which simplifies to 4x1204x - 120. Setting up the inequality for a profit of at least 180180 gives 4x1201804x - 120 \geq 180. Solving for xx gives 4x3004x \geq 300, which simplifies to x75x \geq 75. Therefore, the minimum number of sandwiches that must be sold is 7575.

Step-by-Step Solution

1
Define the variable and write expressions for revenue and cost.
Let xx represent the number of sandwiches sold. Total Revenue = 6.50x6.50x and Total Cost = 120+2.50x120 + 2.50x.
Defining the variable and translating the verbal descriptions of revenue and cost into algebraic expressions is necessary to model the profit.
2
Formulate the net profit expression.
Net Profit = Total Revenue - Total Cost = 6.50x(120+2.50x)=4x1206.50x - (120 + 2.50x) = 4x - 120.
Net profit is calculated by subtracting all fixed and variable costs from the total revenue.
3
Set up and solve the linear inequality.
4x1201804x300x754x - 120 \geq 180 \Rightarrow 4x \geq 300 \Rightarrow x \geq 75.
To find the minimum number of sandwiches needed to reach a target profit of at least 180180, we solve the inequality 4x1201804x - 120 \geq 180 for xx.

Key Concept

Translating a real-world scenario into a linear inequality and solving for the unknown variable.
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