Problem 42 - Algebra Form and Function

Question

You want to build a patio. Builder \(A\) charges \(\$ 3\) a square foot plus a \(\$ 500\) flat fee, and builder \(B\) charges \(\$ 2.50\) a square foot plus a \(\$ 750\) flat fee. For each builder, write an expression relating the cost \(C\) to the area \(s\) square feet of the patio. Which builder is cheaper for a 200 square foot patio? Which is cheaper for a 1000 square foot patio? For what size patio will both builders charge the same?

Step-by-Step Solution

Answer

Answer: For a 200 square foot patio, Builder A is cheaper with a cost of $1100, while Builder B would charge $1250. For a 1000 square foot patio, Builder B is cheaper with a cost of $3250, while Builder A would charge $3500. At a patio size of 500 square feet, both builders would charge the same amount.

Step 1: Write the cost expressions for Builder A and Builder B

Based on the information given, we can write the cost expressions for each builder: Builder A: \(C = 3s + 500\) Builder B: \(C = 2.50s + 750\)

Step 2: Calculate costs for a 200 square foot patio

We'll plug in \(s = 200\) for both cost expressions and compare the results. Builder A: \(C = 3(200) + 500 = 600 + 500 = \$ 1100\) Builder B: \(C = 2.50(200) + 750 = 500 + 750 = \$ 1250\) For a 200 square foot patio, Builder A is cheaper.

Step 3: Calculate costs for a 1000 square foot patio

We'll plug in \(s = 1000\) for both cost expressions and compare the results. Builder A: \(C = 3(1000) + 500 = 3000 + 500 = \$ 3500\) Builder B: \(C = 2.50(1000) + 750 = 2500 + 750 = \$ 3250\) For a 1000 square foot patio, Builder B is cheaper.

Step 4: Determine the size of the patio, for which both builders charge the same

To find the patio size at which both builders charge the same, we'll set the cost expressions equal to each other: \(3s + 500 = 2.50s + 750\) Next, solve for s: \(0.50s = 250\) \(s = 500\) For a 500 square foot patio, both builders will charge the same.

Key Concepts

Linear Equations

Linear equations are mathematical sentences that describe a straight line when graphed.
They have variables, constants, and can easily show relationships between quantities.
In this problem, we use linear equations to model how the cost depends on the area of the patio.For Builder A, the equation is expressed as:

• \( C = 3s + 500 \)

This equation tells us:

• \( 3s \) represents the variable cost, which is the cost per square foot multiplied by the number of square feet.
• The \( 500 \) is a flat fee, a constant that doesn’t change with the area.

Similarly, Builder B's cost is represented by:

• \( C = 2.50s + 750 \)

Here:

• \( 2.50s \) relates to the variable cost per square foot.
• The \( 750 \) is its flat fee.

Understanding this form helps us easily compare costs and find where they intersect.

Problem Solving

Problem-solving in algebra often involves identifying what is known and what needs to be determined.
Here, the task is to find and compare the costs of building a patio for different areas using the given expressions.
This involves simple substitution.To solve effectively:

• Start by substituting the given area values into each builder's cost equation to find the specific cost.
• For example, plugging \( s = 200 \) or \( s = 1000 \) into both equations reveals which builder is more cost-efficient for each size.

Next, you explore when both costs are equal:

• Set the two cost expressions equal, \( 3s + 500 = 2.50s + 750 \), and solve \( s \) for the size where costs match.
• This step requires rearranging the equation to isolate \( s \), showing the logic of linking real-world problems with mathematical operations.

Comparing Costs

Comparing costs is an essential decision-making tool.
Through algebraic expressions, it’s easier to visually and numerically analyze which builder provides the better deal.
For example:

• Calculate costs for both builders for each predefined area, like \( s = 200 \) or \( s = 1000 \), to see immediately which is less expensive.

Finding the size where costs balance:

• We've determined that both builders charge the same for a 500 square foot patio.
• We do this by solving the equality \( 3s + 500 = 2.50s + 750 \).

While comparing, it becomes clear that smaller patios favor Builder A with their lower flat fee,
while larger ones benefit from Builder B's lower cost per square foot.
Knowing these details makes cost analysis in algebra incredibly practical and informative.