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 parabola functions with key points and equations labeled

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. — 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). — Graph of a parabolic function with the vertex labeled

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.