Analysis of Quadratic Functions: Learn It 2

Giselle Miller

Analysis of Quadratic Functions: Learn It 2

Finding the Maximum and Minimum Value of a Quadratic Function

There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.

The vertex of a parabola is the highest (maximum) or lowest (minimum) point, depending on the direction the parabola opens. Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).

Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).

A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased [latex]80[/latex] feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side. Diagram of the garden and the backyard.

Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[/latex]. Then, use the formula to answer: What dimensions should she make her garden to maximize the enclosed area?

Let’s use a diagram to record the given information.

Let [latex]L[/latex] represents the length of the fence and [latex]W[/latex] represents the width of the fence.

We know we have only [latex]80[/latex] feet of fence available, and [latex]L+W+L=80[/latex], or more simply, [latex]2L+W=80[/latex].

This allows us to represent the width, [latex]W[/latex], in terms of [latex]L[/latex].

[latex]W=80 - 2L[/latex]

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.

[latex]A=L \cdot W=L \cdot (80 - 2L)=80L - 2{L}^{2}[/latex].

This formula represents the area of the fence in terms of the variable length [latex]L[/latex].

Thus, the function, written in general form, is[latex]A\left(L\right)=-2{L}^{2}+80L[/latex].

Now, to answer the question: What dimensions should she make her garden to maximize the enclosed area?

The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area.

Based on the function above, we have [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].

Vertex:

[latex]h= -\dfrac{b}{2a} = -\dfrac{80}{2(-2)} = -\dfrac{80}{-4} = 20[/latex]

and

[latex]\begin{align}k&=A\left(20\right) \\&=80\left(20\right)-2{\left(20\right)}^{2}\\&=800 \end{align}[/latex]

Thus, the vertex is [latex](20, 800)[/latex].

Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800). This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.

Interpretation of the vertex: The maximum value of the function is an area of [latex]800[/latex] square feet, which occurs when [latex]L=20[/latex] feet. When the shorter sides are [latex]L = 20[/latex] feet, there is [latex]W = 80-2(20) = 40[/latex] feet of fencing left for the longer side.

So, to maximize the area, she should enclose the garden so the two shorter sides have length [latex]20[/latex] feet and the longer side parallel to the existing fence has length [latex]40[/latex] feet.

The problem we solved above is called a constrained optimization problem. We can optimize our desired outcome given a constraint, which in this case was a limited amount of fencing materials.