A Track And Field Playing Area Is In The Shape Of A Rectangle With Semicircles At Each End. The Inside Perimeter Of The Track Is To Be 1500 Meters. What Should The Dimensions Of The Rectangle Be So That The Area Of The Rectangle Is A Maximum?Dimensions
Introduction
In this article, we will explore a classic optimization problem in mathematics, where we need to find the dimensions of a rectangle with semicircles at each end to maximize its area. The inside perimeter of the track is given as 1500 meters. We will use mathematical techniques to find the optimal dimensions of the rectangle.
Problem Statement
A track and field playing area is in the shape of a rectangle with semicircles at each end. The inside perimeter of the track is to be 1500 meters. We need to find the dimensions of the rectangle so that the area of the rectangle is a maximum.
Mathematical Formulation
Let's denote the length of the rectangle as L and the width as W. The perimeter of the rectangle is given by:
P = 2L + 2W
Since the inside perimeter of the track is 1500 meters, we can set up the following equation:
2L + 2W = 1500
Simplifying the equation, we get:
L + W = 750
Now, let's consider the area of the rectangle, which is given by:
A = LW
We want to maximize the area A subject to the constraint L + W = 750.
Optimization Technique
To solve this optimization problem, we can use the method of Lagrange multipliers. However, in this case, we can use a simpler technique called the "method of substitution".
We can substitute the constraint L + W = 750 into the area equation A = LW to get:
A = (750 - W)W
Expanding the equation, we get:
A = 750W - W^2
Now, we can take the derivative of A with respect to W and set it equal to zero to find the critical point:
dA/dW = 750 - 2W = 0
Solving for W, we get:
W = 375
Now, we can substitute W = 375 back into the constraint L + W = 750 to find the value of L:
L = 750 - W = 750 - 375 = 375
So, the dimensions of the rectangle that maximize the area are L = 375 meters and W = 375 meters.
Conclusion
In this article, we have solved a classic optimization problem in mathematics, where we needed to find the dimensions of a rectangle with semicircles at each end to maximize its area. We used the method of substitution to find the critical point and determined that the dimensions of the rectangle that maximize the area are L = 375 meters and W = 375 meters.
Optimization Results
| Dimension | Value |
|---|---|
| L | 375 meters |
| W | 375 meters |
Discussion
The optimization problem we solved in this article is a classic example of a constrained optimization problem. We used the method of substitution to find the critical point and determined that the dimensions of the rectangle that maximize the area are L = 375 meters and W = 375 meters.
The result we obtained is not surprising, as the rectangle with equal length and width is a square, which is a symmetrical shape. The symmetry of the shape leads to the optimal solution, where the length and width are equal.
Real-World Applications
The optimization problem we solved in this article has many real-world applications in engineering, architecture, and design. For example, in the design of a sports stadium, the dimensions of the playing field need to be optimized to maximize the area while satisfying the constraints of the perimeter and the shape of the field.
In conclusion, the optimization problem we solved in this article is a classic example of a constrained optimization problem, and the result we obtained is not surprising, as the rectangle with equal length and width is a square, which is a symmetrical shape. The symmetry of the shape leads to the optimal solution, where the length and width are equal.
Future Work
In future work, we can explore other optimization problems in mathematics, such as the optimization of a function subject to multiple constraints. We can also use more advanced optimization techniques, such as the method of Lagrange multipliers, to solve more complex optimization problems.
References
- [1] Boyd, S., & Vandenberghe, L. (2004). Convex optimization. Cambridge University Press.
- [2] Bertsekas, D. P. (1999). Nonlinear programming. Athena Scientific.
- [3] Luenberger, D. G. (1984). Linear and nonlinear programming. Addison-Wesley.
Appendix
The following is a list of the mathematical formulas used in this article:
P = 2L + 2W(perimeter of the rectangle)L + W = 750(constraint)A = LW(area of the rectangle)A = (750 - W)W(substituted area equation)dA/dW = 750 - 2W = 0(derivative of the area equation)W = 375(critical point)L = 750 - W = 375(value of L)
Introduction
In our previous article, we explored a classic optimization problem in mathematics, where we needed to find the dimensions of a rectangle with semicircles at each end to maximize its area. The inside perimeter of the track is given as 1500 meters. We used mathematical techniques to find the optimal dimensions of the rectangle.
In this article, we will answer some frequently asked questions (FAQs) related to the optimization problem.
Q&A
Q: What is the optimization problem about?
A: The optimization problem is about finding the dimensions of a rectangle with semicircles at each end to maximize its area. The inside perimeter of the track is given as 1500 meters.
Q: What is the constraint in the optimization problem?
A: The constraint in the optimization problem is that the inside perimeter of the track is 1500 meters, which can be expressed as L + W = 750, where L is the length of the rectangle and W is the width.
Q: How do we find the optimal dimensions of the rectangle?
A: We use the method of substitution to find the critical point and determine the optimal dimensions of the rectangle.
Q: What is the optimal solution to the optimization problem?
A: The optimal solution to the optimization problem is that the dimensions of the rectangle that maximize the area are L = 375 meters and W = 375 meters.
Q: Why is the optimal solution a square?
A: The optimal solution is a square because the rectangle with equal length and width is a symmetrical shape. The symmetry of the shape leads to the optimal solution, where the length and width are equal.
Q: What are some real-world applications of the optimization problem?
A: The optimization problem has many real-world applications in engineering, architecture, and design. For example, in the design of a sports stadium, the dimensions of the playing field need to be optimized to maximize the area while satisfying the constraints of the perimeter and the shape of the field.
Q: Can we use other optimization techniques to solve the problem?
A: Yes, we can use other optimization techniques, such as the method of Lagrange multipliers, to solve the problem. However, the method of substitution is a simpler and more straightforward approach.
Q: What are some common mistakes to avoid when solving optimization problems?
A: Some common mistakes to avoid when solving optimization problems include:
- Not clearly defining the objective function and the constraints
- Not using the correct optimization technique for the problem
- Not checking the validity of the solution
- Not considering the physical constraints of the problem
Q: How can we apply the optimization problem to real-world scenarios?
A: We can apply the optimization problem to real-world scenarios by:
- Identifying the objective function and the constraints of the problem
- Using the correct optimization technique to find the optimal solution
- Checking the validity of the solution
- Considering the physical constraints of the problem
Conclusion
In this article, we have answered some frequently asked questions (FAQs) related to the optimization problem. We have discussed the optimization problem, the constraint, the optimal solution, and some real-world applications of the problem. We have also provided some tips on how to apply the optimization problem to real-world scenarios.
Optimization Results
| Dimension | Value |
|---|---|
| L | 375 meters |
| W | 375 meters |
Discussion
The optimization problem we solved in this article is a classic example of a constrained optimization problem. We used the method of substitution to find the critical point and determined that the dimensions of the rectangle that maximize the area are L = 375 meters and W = 375 meters.
The result we obtained is not surprising, as the rectangle with equal length and width is a square, which is a symmetrical shape. The symmetry of the shape leads to the optimal solution, where the length and width are equal.
Real-World Applications
The optimization problem we solved in this article has many real-world applications in engineering, architecture, and design. For example, in the design of a sports stadium, the dimensions of the playing field need to be optimized to maximize the area while satisfying the constraints of the perimeter and the shape of the field.
Future Work
In future work, we can explore other optimization problems in mathematics, such as the optimization of a function subject to multiple constraints. We can also use more advanced optimization techniques, such as the method of Lagrange multipliers, to solve more complex optimization problems.
References
- [1] Boyd, S., & Vandenberghe, L. (2004). Convex optimization. Cambridge University Press.
- [2] Bertsekas, D. P. (1999). Nonlinear programming. Athena Scientific.
- [3] Luenberger, D. G. (1984). Linear and nonlinear programming. Addison-Wesley.
Appendix
The following is a list of the mathematical formulas used in this article:
P = 2L + 2W(perimeter of the rectangle)L + W = 750(constraint)A = LW(area of the rectangle)A = (750 - W)W(substituted area equation)dA/dW = 750 - 2W = 0(derivative of the area equation)W = 375(critical point)L = 750 - W = 375(value of L)
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