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Chapter 4: Problem 2

a. Identify the output variable to be optimized and the input variable(s). b. Sketch and label a diagram. c. Write a model for the output variable in terms of one input variable. d. Answer the question posed. A mason has enough brick to build a 46 foot wall. The homeowners want to use the wall to enclose an outdoor patio. The patio will be along the side of their house and will include a 4 -foot opening for a door. What dimensions will maximize the area of the patio?

Short Answer

Expert verified
The dimensions that maximize the patio area are 10.5 feet by 21 feet.

Step by step solution

01

Identify Variables

The problem involves maximizing the area of a rectangular patio enclosed by a wall against a house. The output variable that needs to be optimized is the area of the patio. The input variables are the dimensions of the patio: width (\(w\)) and length (\(l\)).
02

Sketch the Diagram

Visualize the situation by sketching a rectangle with one side along the house. Label the side parallel to the house as \(l\) and perpendicular to the house as \(w\). The wall comprises the sum of \(w\), \(w\), and \(l\) with a 4-foot gap for a door.
03

Formulate the Equation

The masonry constraint is that the total length of the wall except the 4-foot door is 46 feet. Hence, the equation is \(2w + l + 4 = 46\). Simplify this to \(2w + l = 42\). The area \(A\) to be maximized is given by the product of width and length: \(A = w \times l\).
04

Express Length in Terms of Width

Rearrange the wall constraint equation \(2w + l = 42\) to express \(l\) in terms of \(w\): \(l = 42 - 2w\).
05

Construct the Area Function

Substitute \(l = 42 - 2w\) back into the area equation to get \(A(w) = w(42 - 2w)\). Simplify to \(A(w) = 42w - 2w^2\).
06

Differentiate to Find Critical Points

Differentiate \(A(w) = 42w - 2w^2\) to find the critical points: \(A'(w) = 42 - 4w\). Set this equal to zero: \(42 - 4w = 0\) and solve for \(w\): \(w = 10.5\).
07

Determine Maximum Area

Substitute \(w = 10.5\) into \(l = 42 - 2w\) to find \(l\). So, \(l = 42 - 21 = 21\). Calculate the area \(A = 10.5 \times 21 = 220.5\) square feet.
08

Verify and Conclude

Check that the walls add up correctly: \(2 \times 10.5 + 21 = 42\). The dimensions that maximize the area of the patio are width 10.5 feet and length 21 feet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area Maximization
Maximizing the area of the patio is the primary goal of this exercise. This involves finding the optimal length and width dimensions that will yield the largest possible rectangular area. When solving an area maximization problem, especially in calculus, we work to determine the largest possible value for the area under given constraints, such as fixed perimeter or boundary lengths. For the patio problem, the constraint is the total length of the wall (46 feet, discounting the 4-foot opening for the door, which leaves 42 feet). By rearranging and optimizing these dimensions using calculus, we can find the maximum area that the patio can occupy.
Rectangular Dimensions
In this scenario, the patio is envisioned as a rectangular area that needs to fit against a house. To optimize the space mathematically, understanding how these rectangular dimensions contribute to area calculation is crucial.Key dimensions include:- Width (\(w\)): how far the patio extends away from the house.- Length (\(l\)): how long the patio runs parallel to the house.The relationship given is described by the equation \(2w + l = 42\), simplifying the structural limit. The optimization seeks to adjust \(w\) and \(l\) to use this full extent while maximizing area.
Differentiation
Differentiation is a pivotal tool in optimization problems. It enables us to find the rate at which the area function changes with respect to the dimension \(w\), and identify points where this rate is zero, known as critical points. This is done by setting the derivative of the area function to zero.Using the area function \(A(w) = 42w - 2w^2\), we differentiating it to find \(A'(w) = 42 - 4w\). Solving \(A'(w) = 0\) will point us directly to the \(w\) value that may maximize the patio's area.Thus, differentiation becomes the pathway through which we ascertain the most efficient dimensions of the patio.
Critical Points
Critical points are values of the input variable where the derivative equals zero or does not exist. They indicate potential locations for maximum or minimum values of the function being examined. In this exercise, these points guide us in identifying the maximum area the patio can attain.For the given problem, we found the critical point by solving \(42 - 4w = 0\), yielding \(w = 10.5\). This \(w\) determines the width that, when used along with the corresponding length \(l = 21\), results in the maximum area of \(220.5\) square feet.Evaluating and checking these critical points ensures the solution is correct and the dimensions are optimal for enclosing the patio area against the wall.

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