/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A farmer plans to fence a rectan... [FREE SOLUTION] | 91Ó°ÊÓ

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A farmer plans to fence a rectangular pasture adjacent to a river. The pasture must contain 180,000 square meters in order to provide enough grass for the herd. What dimensions would require the least amount of fencing if no fencing is needed along the river?

Short Answer

Expert verified
The dimensions that would require the least amount of fencing are approximately length = 598m, width = 301m.

Step by step solution

01

Express the length in terms of width

We know that the area \( A = length \times width \). So, we can express the length in terms of the width as \( Length = \frac{A}{width} = \frac{180,000}{width} \). So this is our equation [1].
02

Express the Perimeter in terms of width

The Perimeter is given by \( P = length + 2 \times width \). Replace the length from equation [1] to express perimeter in terms of width. So, \( P = \frac{180,000}{width} + 2 \times width \).
03

Find the derivative of the Perimeter

Next, let's find the derivative of the perimeter with respect to the width. The derivative of \( P = \frac{180,000}{width} + 2 \times width \) gives us \( P' = -\frac{180,000}{width^2} + 2 \).
04

Find the width that minimizes the Perimeter

Set the derivative equal to zero and solve to find the width. So, \( -\frac{180,000}{width^2} + 2 = 0 \). Solving this gives \( width = \sqrt{\frac{180,000}{2}} \approx 301m \).
05

Find the length

Using the value of width in equation [1], we get \( length = \frac{180,000}{301} \approx 598m \).

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that assists in understanding change and motion. It involves using derivatives and integrals to study how quantities change. In optimization problems like the one in our example, calculus helps us determine the maximum or minimum values for a given situation.
In this exercise, using calculus is a key part to discover the dimensions of the rectangular pasture that minimize the fencing needed. By creating equations for the area and perimeter, we use calculus to derive a formula that can be optimized. It provides a solid mathematical approach to solving real-world problems efficiently, ensuring resources are conserved while meeting operational constraints.
Derivatives
Derivatives are a foundational concept in calculus that measure the rate at which a quantity changes. In this exercise, they are used to find the slope of the perimeter function with respect to width. This is crucial because the derivative allows us to locate the minimum point in the function.
By taking the derivative of the perimeter formula, which is expressed as a function of width, we end up with a new function. The critical part is setting this derivative equal to zero, helping us identify where the rate of change switches from decreasing to increasing. This is the width where the perimeter is minimized, providing an optimal solution without excessive material usage.
Geometry
Understanding geometry is key in this problem as it involves shapes and their properties. Here, we are dealing with a rectangle adjacent to a river, influencing our calculations. Geometry provides the tools to express the relationships between different parts of the shape, such as the length, width, and area.
For instance, calculating the area of the pasture involves multiplying the length by the width, a basic geometric property of rectangles. Equally, the geometric notion that fencing isn't required along the river allows us to create a more efficient equation for perimeter, focusing solely on the fenced sides. Geometry thus lays the foundation to set up correct equations and relationships necessary for solving the problem.
Perimeter Minimization
Perimeter minimization in this problem is about finding the dimensions that reduce the amount of fencing required. This concept is practical in scenarios where one side doesn't require fencing, such as the side adjacent to the river in our case.
The minimization process involves turning the physical situation into a mathematical one, identifying an expression for the perimeter that we can work with. With the derivative, we explore how the perimeter changes with respect to width. Setting the change to zero highlights the optimal dimensions, straightforwardly showing how a calculated approach can conserve resources in real-life applications, like fencing a farm enclosure efficiently.

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Most popular questions from this chapter

Prove Darboux's Theorem: Let \(f\) be differentiable on the closed interval \([a, b]\) such that \(f^{\prime}(a)=y_{1}\) and \(f^{\prime}(b)=y_{2} .\) If \(d\) lies between \(y_{1}\) and \(y_{2}\), then there exists \(c\) in \((a, b)\) such that \(f^{\prime}(c)=d\)

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Consider the functions \(f(x)=\frac{1}{2} x^{2}\) and \(g(x)=\frac{1}{16} x^{4}-\frac{1}{2} x^{2}\) on the domain \([0,4]\). (a) Use a graphing utility to graph the functions on the specified domain. (b) Write the vertical distance \(d\) between the functions as a function of \(x\) and use calculus to find the value of \(x\) for which \(d\) is maximum. (c) Find the equations of the tangent lines to the graphs of \(f\) and \(g\) at the critical number found in part (b). Graph the tangent lines. What is the relationship between the lines? (d) Make a conjecture about the relationship between tangent lines to the graphs of two functions at the value of \(x\) at which the vertical distance between the functions is greatest, and prove your conjecture.

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