Question

An animal shelter can house only cats and dogs. Each dog requires 2 cups of food and 3 treats a day, while each cat requires 1 cup of food a day and 2 treats a day. If the shelter has available a total of 400 cups of food and 500 treats a day, what expressions portray the full scope of the number of \(c\) cats and dogs the shelter could potentially house? (A) \(2 \mathrm{~d}-\mathrm{c} \leq 400\) and \(3 \mathrm{~d}+\mathrm{c}<500\) (B) \(2 d+c \leq 400\) and \(3 d+2 c \leq 500\) (C) \(4 \mathrm{~d}+\mathrm{c}<400\) and \(\mathrm{d}+\mathrm{c}<500\) (D) \(2 \mathrm{~d}+2 \mathrm{c} \leq 400\) and \(2 \mathrm{~d}+3 \mathrm{c} \leq 500\)

Short answer

Option (B): \(2d + c \leq 400\) and \(3d + 2c \leq 500\).

Step-by-step solution

Identify Constraints for Food

Identify how much food is needed by each cat and dog. Each dog requires 2 cups of food per day and each cat requires 1 cup of food per day. The total amount of food available is 400 cups. Therefore, the inequality representing this constraint is: \[2d + c \leq 400\]

Identify Constraints for Treats

Next, determine how many treats are needed by each cat and dog. Each dog requires 3 treats per day and each cat requires 2 treats per day. The total number of treats available is 500. This gives us the inequality: \[3d + 2c \leq 500\]

Match Constraints to Given Options

Now compare the inequalities derived from the constraints to the provided answer options. The inequalities \(2d + c \leq 400\) and \(3d + 2c \leq 500\) match exactly with option (B).

Key concepts

Inequalities in Word Problems

When we encounter word problems, especially those dealing with limits or constraints, inequalities are a powerful tool. They help us represent conditions that solutions must satisfy. In this exercise, the problem states that the animal shelter has fixed amounts of food and treats.
This scenario naturally leads to using inequalities to express limits. The food inequality

• The inequality \(2d + c \leq 400\) shows that the total cups of food needed for \(d\) dogs and \(c\) cats cannot exceed 400 cups.
• Similarly, the treats inequality \(3d + 2c \leq 500\) ensures that the total treats consumed by the animals must not surpass 500 treats.

By using inequalities, we model real-world restrictions mathematically, allowing us to understand, analyze, and optimize decisions effectively.

Algebraic Expressions

Algebra serves as the language of mathematics, where we use expressions to succinctly represent complex ideas.
In this problem, algebraic expressions are critical to defining how variables relate within the animal shelter's constraints. Let's break it down:

• The term \(2d + c\) represents the total amount of food consumed by all animals. Here, \(d\) is the number of dogs, each consuming 2 cups of food, and \(c\) is the number of cats, each consuming 1 cup.
• Meanwhile, \(3d + 2c\) captures the total treats required each day. Dogs eat 3 treats, cats eat 2.

The brilliance of algebra is that it allows us to easily substitute different quantities, thereby exploring potential solutions while abiding by the constraints.

Animal Shelter Optimization Problem

The ultimate aim in this word problem is to optimize how many cats and dogs the shelter can house, given limited resources like food and treats. This is an optimization problem, where the goal is to maximize or effectively use resources without exceeding availability.
Real-world applications of such problems include:

• Ensuring the shelter does not exceed 400 cups of food, modeled by \(2d + c \leq 400\).
• Making sure treats do not go over 500, addressed by \(3d + 2c \leq 500\).

Finding the combination of \(d\) and \(c\) that fits these conditions most economically helps achieve an optimal solution. This approach is often used in resource management, logistics, and planning to make informed decisions