Darius Is Making Picture Frames. He Starts With A Cardboard Piece That Is 12 Inches Long On Each Side. He Then Cuts Out A Square Hole, Where The Size Of The Square Depends On The Picture He Is Framing.Let S S S Be The Length (in Inches) Of One

Introduction

In this article, we will delve into a problem presented by Darius, who is making picture frames. The problem revolves around a cardboard piece that is 12 inches long on each side, and a square hole that is cut out to accommodate the picture. The size of the square hole depends on the picture being framed. We will explore the mathematical concepts involved in this problem and provide a step-by-step solution.

The Problem

Let $s$ be the length (in inches) of one side of the square hole. The area of the square hole is given by $s^2$. The area of the entire cardboard piece is $12^2 = 144$ square inches. Since the square hole is cut out from the cardboard piece, the remaining area is $144 - s^2$.

The Question

The question is: what is the maximum value of $s$ such that the remaining area is still positive?

Mathematical Analysis

To solve this problem, we need to find the maximum value of $s$ such that $144 - s^2 > 0$. This inequality can be rewritten as $s^2 < 144$. Taking the square root of both sides, we get $|s| < 12$. Since $s$ represents the length of one side of the square hole, it must be positive. Therefore, we can write $0 < s < 12$.

Graphical Representation

To visualize the problem, we can graph the function $f(s) = 144 - s^2$. The graph of this function is a downward-facing parabola that opens downwards. The maximum value of $s$ occurs when the graph intersects the x-axis, which happens when $s = 12$. However, since $s$ must be positive, the maximum value of $s$ is actually $s = 12 - \epsilon$, where $\epsilon$ is an arbitrarily small positive number.

Conclusion

In conclusion, the maximum value of $s$ such that the remaining area is still positive is $s = 12 - \epsilon$. This means that Darius can cut out a square hole with a side length of up to $12 - \epsilon$ inches without reducing the remaining area to zero.

Step-by-Step Solution

Here is a step-by-step solution to the problem:

1. Write down the inequality $144 - s^2 > 0$.
2. Rewrite the inequality as $s^2 < 144$.
3. Take the square root of both sides to get $|s| < 12$.
4. Since $s$ must be positive, write $0 < s < 12$.
5. Graph the function $f(s) = 144 - s^2$ to visualize the problem.
6. Find the maximum value of $s$ by finding the x-intercept of the graph.
7. Since $s$ must be positive, the maximum value of $s$ is $s = 12 - \epsilon$.

Real-World Applications

This problem has real-world applications in various fields, such as:

• Framing: The problem is relevant to framing pictures, where the size of the square hole depends on the picture being framed.
• Design: The problem is also relevant to design, where the size of the square hole depends on the design of the picture frame.
• Mathematics: The problem is a classic example of a mathematical optimization problem, where the goal is to maximize the size of the square hole while keeping the remaining area positive.

Conclusion

Introduction

In our previous article, we explored the problem presented by Darius, who is making picture frames. The problem revolves around a cardboard piece that is 12 inches long on each side, and a square hole that is cut out to accommodate the picture. We provided a step-by-step solution to the problem and discussed its real-world applications. In this article, we will answer some frequently asked questions related to the problem.

Q&A Session

Q: What is the maximum value of s such that the remaining area is still positive?

A: The maximum value of s is 12 - ε, where ε is an arbitrarily small positive number.

Q: Why can't s be equal to 12?

A: Since s represents the length of one side of the square hole, it must be positive. Therefore, s cannot be equal to 12, but rather 12 - ε.

Q: What is the remaining area when s is equal to 12 - ε?

A: The remaining area is 144 - (12 - ε)^2, which approaches 0 as ε approaches 0.

Q: Can s be greater than 12?

A: No, s cannot be greater than 12, since the remaining area would be negative.

Q: What is the minimum value of s such that the remaining area is still positive?

A: The minimum value of s is 0, since the remaining area is still positive when s is equal to 0.

Q: How does the size of the square hole affect the remaining area?

A: The size of the square hole affects the remaining area in a non-linear way. As the size of the square hole increases, the remaining area decreases.

Q: Can the size of the square hole be greater than 12?

A: No, the size of the square hole cannot be greater than 12, since the remaining area would be negative.

Q: What is the relationship between the size of the square hole and the remaining area?

A: The size of the square hole and the remaining area are inversely proportional. As the size of the square hole increases, the remaining area decreases.

Real-World Applications

The problem presented by Darius has real-world applications in various fields, such as:

• Framing: The problem is relevant to framing pictures, where the size of the square hole depends on the picture being framed.
• Design: The problem is also relevant to design, where the size of the square hole depends on the design of the picture frame.
• Mathematics: The problem is a classic example of a mathematical optimization problem, where the goal is to maximize the size of the square hole while keeping the remaining area positive.

Conclusion

In conclusion, the problem presented by Darius is a classic example of a mathematical optimization problem. The solution involves finding the maximum value of s such that the remaining area is still positive. The problem has real-world applications in various fields, such as framing, design, and mathematics. We hope that this Q&A session has provided a better understanding of the problem and its solutions.

Additional Resources

For further reading and exploration, we recommend the following resources:

• Mathematical Optimization: A comprehensive resource on mathematical optimization, including problems and solutions.
• Framing and Design: A resource on framing and design, including tips and techniques for creating picture frames.
• Mathematics: A resource on mathematics, including problems and solutions, as well as tutorials and examples.

Conclusion

In conclusion, the problem presented by Darius is a classic example of a mathematical optimization problem. The solution involves finding the maximum value of s such that the remaining area is still positive. The problem has real-world applications in various fields, such as framing, design, and mathematics. We hope that this Q&A session has provided a better understanding of the problem and its solutions.