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A lifeguard needs to rope off a rectangular swimming area in front of Long Lake Beach, using 2100 yd of rope and floats. What dimensions of the rectangle will maximize the area? What is the maximum area? (Note that the shoreline is one side of the rectangle and thus does not need rope.) A(x) = Length of shorter side of the rectangle region: Length of longer side of the rectangle region: Maximum area of the rectangular region:

A lifeguard needs to rope off a rectangular swimming area in front of Long Lake Beach, using 2100 yd of rope and floats. What dimensions of the rectangle will maximize the area? What is the maximum area? (Note that the shoreline is one side of the rectangle and thus does not need rope.) A(x) = Length of shorter side of the rectangle region: Length of longer side of the rectangle region: Maximum area of the rectangular region:

College Algebra (MindTap Course List)
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter4: Polynomial And Rational Functions
Section4.1: Quadratic Functions
Problem 61E: Police investigations A police officer seals off the scene of an accident using a roll of yellow...
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### Optimizing a Rectangular Swimming Area at Long Lake Beach A lifeguard needs to rope off a rectangular swimming area in front of Long Lake Beach, using 2100 yards of rope and floats. The question is to determine the dimensions of the rectangle that will maximize the area. Additionally, we need to find the maximum area that can be achieved. Note that the shoreline is one side of the rectangle, and thus does not require rope. Let's proceed with the following information: 1. **Function for Area:** - `A(x) = __________` 2. **Lengths of the Rectangle's Sides:** - Length of the shorter side of the rectangle region: `__________` - Length of the longer side of the rectangle region: `__________` 3. **Maximum Area:** - Maximum area of the rectangular region: `__________` Here's the mathematical setup for the problem: **Given:** - Total length of rope available: 2100 yards - Shoreline acts as one side, so rope is needed for the other three sides. **Formulation:** - Let `x` be the length of the shorter side perpendicular to the shoreline. - The other side (parallel to the shoreline) will be `y`. **Constraints:** - Rope used: `2x + y = 2100` To maximize the rectangular area `A`, the area can be expressed as: - `A = x * y` Using the constraint equation, replace `y`: - `y = 2100 - 2x` Thus, the area function in terms of `x` is: - `A(x) = x * (2100 - 2x)` Expanding this: - `A(x) = 2100x - 2x^2` To find the value of `x` that maximizes the area, take the derivative of `A(x)` and set it to zero: - `A'(x) = 2100 - 4x = 0` - Solving for `x`, we get: `x = 525` Now, substitute `x = 525` back into the constraint equation to find `y`: - `y = 2100 - 2*525 = 1050` **Recording the Values:** - `A(x) = 2100x - 2x^2` - Length of shorter side
Transcribed Image Text:### Optimizing a Rectangular Swimming Area at Long Lake Beach A lifeguard needs to rope off a rectangular swimming area in front of Long Lake Beach, using 2100 yards of rope and floats. The question is to determine the dimensions of the rectangle that will maximize the area. Additionally, we need to find the maximum area that can be achieved. Note that the shoreline is one side of the rectangle, and thus does not require rope. Let's proceed with the following information: 1. **Function for Area:** - `A(x) = __________` 2. **Lengths of the Rectangle's Sides:** - Length of the shorter side of the rectangle region: `__________` - Length of the longer side of the rectangle region: `__________` 3. **Maximum Area:** - Maximum area of the rectangular region: `__________` Here's the mathematical setup for the problem: **Given:** - Total length of rope available: 2100 yards - Shoreline acts as one side, so rope is needed for the other three sides. **Formulation:** - Let `x` be the length of the shorter side perpendicular to the shoreline. - The other side (parallel to the shoreline) will be `y`. **Constraints:** - Rope used: `2x + y = 2100` To maximize the rectangular area `A`, the area can be expressed as: - `A = x * y` Using the constraint equation, replace `y`: - `y = 2100 - 2x` Thus, the area function in terms of `x` is: - `A(x) = x * (2100 - 2x)` Expanding this: - `A(x) = 2100x - 2x^2` To find the value of `x` that maximizes the area, take the derivative of `A(x)` and set it to zero: - `A'(x) = 2100 - 4x = 0` - Solving for `x`, we get: `x = 525` Now, substitute `x = 525` back into the constraint equation to find `y`: - `y = 2100 - 2*525 = 1050` **Recording the Values:** - `A(x) = 2100x - 2x^2` - Length of shorter side
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