Example 1

A backyard farmer wants to enclose a rectangular space for a new garden. She has purchased 80 feet of wire fencing to enclose 3 sides, and will put the 4th side against the backyard fence. Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length L. In a scenario like this involving geometry, it is often helpful to draw a picture. It might also be helpful to introduce a temporary variable, W, to represent the side of fencing parallel to the 4th side or backyard fence. Since we know we only have 80 feet of fence available, we know that L + W + L = 80, or more simply, 2L + W = 80. This allows us to represent the width, W, in terms of L: W = 80 – 2L Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so A = LW = L(80 – 2L) A(L) = 80L – 2L² This formula represents the area of the fence in terms of the variable length L.

Example 2

Returning to our backyard farmer from the beginning of the section, what dimensions should she make her garden to maximize the enclosed area? Earlier we determined the area she could enclose with 80 feet of fencing on three sides was given by the equation A(L) = 80L – 2L². Notice that quadratic has been vertically reflected, since the coefficient on the squared term is negative, so the graph will open downwards, and the vertex will be a maximum value for the area. In finding the vertex, we take care since the equation is not written in standard polynomial form with decreasing powers. But we know that a is the coefficient on the squared term, so a = -2, b = 80, and c = 0. Finding the vertex: [latex-display]\displaystyle{h}=-\frac{{80}}{{{2}{(-{2})}}}={20},{k}={A}{({20})}={80}{({20})}-{2}{({20})}^{{2}}={800}[/latex-display] The maximum value of the function is an area of 800 square feet, which occurs when L = 20 feet. When the shorter sides are 20 feet, that leaves 40 feet of fencing for the longer side. To maximize the area, she should enclose the garden so the two shorter sides have length 20 feet, and the longer side parallel to the existing fence has length 40 feet.

Example 3

A ball is thrown upwards from the top of a 40 foot high building at a speed of 80 feet per second. The ball's height above ground can be modeled by the equation H(t) = –16t2 + 80t + 40. What is the maximum height of the ball? When does the ball hit the ground? To find the maximum height of the ball, we would need to know the vertex of the quadratic. [latex-display]\displaystyle{h}=-\frac{{80}}{{{2}{(-{16})}}}=\frac{{80}}{{32}}=\frac{{5}}{{2}},{k}={H}{(\frac{{5}}{{2}})}=-{16}{(\frac{{5}}{{2}})}^{{2}}+{80}{(\frac{{5}}{{2}})}+{40}={140}[/latex-display] The ball reaches a maximum height of 140 feet after 2.5 seconds. To find when the ball hits the ground, we need to determine when the height is zero—when H(t) = 0. While we could do this using the transformation form of the quadratic, we can also use the quadratic formula: [latex-display]\displaystyle{t}=\frac{{-{80}\pm\sqrt{{{80}^{{2}}-{4}{(-{16})}{({40})}}}}}{{{2}{(-{16})}}}=\frac{{-{80}sqrt{{{8960}}}}}{{-{32}}}[/latex-display] Since the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions: [latex-display]\displaystyle{t}=\frac{{-{80}-sqrt{{8960}}}}{{-{{32}}}}approx{5.458}[/latex-display] The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds.

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David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, " Chapter 3: Polynomial and Rational Functions," licensed under a CC BY-SA 3.0 license.