Problem 39 - Elementary and Intermediate Algebra

Question

Solve each problem by writing a variation model. The number of days that a given number of bushels of corn will last when feeding cattle varies inversely as the number of animals. If \(x\) bushels will feed 25 cows for 10 days, how long will the feed last for 10 cows?

Step-by-Step Solution

Answer

The feed will last 25 days for 10 cows.

Step 1: Understand the Problem

The exercise states that the number of days ( \(d\)) the feed will last varies inversely as the number of cows (\(a\)). This means the relationship can be modeled as \(d = \frac{k}{a}\), where \(k\) is a constant.

Step 2: Find the Constant of Variation

We are given that \(x\) bushels will last 25 cows for 10 days. Substitute into the inverse variation equation: \(10 = \frac{k}{25}\). Multiply both sides by 25 to solve for \(k\): \(k = 25 \times 10 = 250\).

Step 3: Set Up the New Problem with Known Value

Use the constant \(k = 250\) to find how long the feed will last for 10 cows. Substitute into the inverse variation equation: \(d = \frac{250}{10}\).

Step 4: Solve for the New Number of Days

Simplify \(d = \frac{250}{10} = 25\). So, the feed will last 25 days for 10 cows.

Key Concepts

Algebra

Algebra is a branch of mathematics that deals with symbols and rules for manipulating those symbols. In this exercise, we use algebraic expressions and equations to model real-world situations, such as determining how long a supply lasts given certain conditions. The algebraic relationship in this problem is an inverse variation. This means one quantity increases when another decreases. Knowing this allows us to form an equation: \[ d = \frac{k}{a} \]Here, \(d\) is the number of days, \(a\) represents the number of animals, and \(k\) is the constant of variation. Algebra helps us manipulate these equations to find unknown values, like the number of days the feed will last under a different number of cows, using given information.

Constant of Variation

In the context of variation problems, the constant of variation \(k\) is a crucial element. It represents a fixed number that relates two variables in an inverse or direct relationship. In our exercise, the constant of variation links the number of days the corn lasts (\(d\)) with the number of cows (\(a\)).To find \(k\), we use the known quantities from the problem: 10 days of feed for 25 cows. By substituting these into the inverse variation equation: \[ 10 = \frac{k}{25} \]We isolate \(k\) by multiplying both sides by 25, giving us \(k = 250\). This constant remains the same regardless of changes in \(d\) and \(a\), simplifying calculations when the number of cows varies.

Variation Models

Variation models help us describe how one quantity changes in relation to another. These models come in two major types: direct and inverse. In direct variation, as one quantity increases, so does the other; whereas, in inverse variation, as one quantity increases, the other decreases.In our exercise, we use an inverse variation model because the number of animals negatively impacts how long the food supply lasts. With an inverse variation:\[ d = \frac{k}{a} \]We substitute the known values and the constant previously calculated to solve new scenarios quickly. For example, using \(k = 250\), and \(10\) cows, we find that the feed will last for 25 days:\[ d = \frac{250}{10} = 25 \]These models simplify problem-solving, allowing us to adapt known scenarios to new conditions with ease.