(10.3.1) – Solve application problems involving quadratic functions

[latex]\large \begin{array}{c} h=-\frac{80}{2\left(-16\right)} \text{ }=\frac{80}{32}\hfill \\ \text{ }=\frac{5}{2}\hfill \\ \text{ }=2.5\hfill \end{array}[/latex]

The ball reaches a maximum height after 2.5 seconds.

b. To find the maximum height, find the y coordinate of the vertex of the parabola.

[latex]\large \begin{array}{c}k=H\left(-\frac{b}{2a}\right)\hfill \\ \text{ }=H\left(2.5\right)\hfill \\ \text{ }=-16{\left(2.5\right)}^{2}+80\left(2.5\right)+40\hfill \\ \text{ }=140\hfill \end{array}[/latex]

The ball reaches a maximum height of 140 feet.

c. To find when the ball hits the ground, we need to determine when the height is zero, [latex]H\left(t\right)=0[/latex].

We use the quadratic formula.

[latex]\large \begin{array}{c} t=\frac{-80\pm \sqrt{{80}^{2}-4\left(-16\right)\left(40\right)}}{2\left(-16\right)}\hfill \\ \text{ }=\frac{-80\pm \sqrt{8960}}{-32}\hfill \end{array}[/latex]

Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.

[latex]\large \begin{array}{c}t=\frac{-80-\sqrt{8960}}{-32}\approx 5.458\hfill & \text{or}\hfill & t=\frac{-80+\sqrt{8960}}{-32}\approx -0.458\hfill \end{array}[/latex]

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.

We are trying to find the minimum of the product [latex]P=xy[/latex] of two numbers, such that their difference is 100: [latex]y-x=100[/latex]. First, we rewrite one variable in terms of the other:

[latex]y-x=100 \rightarrow y=100+x[/latex]

Next, we plug in the above relationship between the variables into the first equation:

[latex]P=xy=x(100+x) = 100x+x^2 = x^2+100x[/latex]

As a result, we get a quadratic function [latex]P(x)=x^2+100x[/latex]. The graph of this quadratic function opens upwards, and its vertex is the minimum, So if we find the vertex of this parabola, we will find the minimum product. The vertex is:

[latex]\displaystyle \displaystyle \left(-\frac{b}{2a}, P\left(-\frac{b}{2a}\right)\right) = \left(-\frac{(100)}{2(1)}, P\left(-\frac{(100)}{2(1)}\right)\right) = (-50,-2,500)[/latex]

Thus the minimum of the parabola occurs at [latex]x=-50[/latex], and is [latex]-2,500[/latex]. So one of the numbers is [latex]x=-50[/latex], the other we obtain by plugging in [latex]x=-50[/latex] in to [latex]y-x=100[/latex]:

[latex]\begin{array}{cc}y-(-50)&=&100 \\ y+50 &=& 100 \\ y &=& 50\end{array}[/latex]

Answer

[latex]x=-50[/latex] and [latex]y=50[/latex]

Let’s use a diagram such as the one above to record the given information. It is also helpful to introduce a temporary variable, W, to represent the width of the garden and the length of the fence section parallel to the backyard fence.

1)  We know we have only 80 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, so

[latex]\begin{array}{l}\text{ }A&=&LW=L\left(80 - 2L\right)\hfill \\ A\left(L\right)&=&80L - 2{L}^{2}\hfill \end{array}[/latex]

This formula represents the area of the fence in terms of the variable length [latex]L[/latex]. The function, written in general form, is

[latex]A\left(L\right)=-2{L}^{2}+80L[/latex].

2) The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since [latex]a[/latex] is the coefficient of the squared term, [latex]a=-2,b=80[/latex], and [latex]c=0[/latex].

To find the vertex:

[latex]\large \begin{array}{l}h=-\frac{80}{2\left(-2\right)}\hfill & \hfill & \hfill & \hfill & k=A\left(20\right)\hfill \\ \text{ }=20\hfill & \hfill & \text{and}\hfill & \hfill & \text{ }=80\left(20\right)-2{\left(20\right)}^{2}\hfill \\ \hfill & \hfill & \hfill & \hfill & \text{ }=800\hfill \end{array}[/latex]

The maximum value of the function is an area of 800 square feet, which occurs when [latex]L=20[/latex] feet. When the shorter sides are 20 feet, there is 40 feet of fencing left 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.

Analysis of the Solution

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.

Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, [latex]p[/latex] for price per subscription and [latex]Q[/latex] for quantity, giving us the equation [latex]\text{Revenue}=pQ[/latex].

Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently [latex]p=30[/latex] and [latex]Q=84,000[/latex]. We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, [latex]p=32[/latex] and [latex]Q=79,000[/latex]. From this we can find a linear equation relating the two quantities. The slope will be

[latex]\large \begin{array}{c}m=\frac{79,000 - 84,000}{32 - 30}\hfill \\ \text{ }=\frac{-5,000}{2}\hfill \\ \text{ }=-2,500\hfill \end{array}[/latex]

This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the y-intercept.

[latex]\begin{array}{c}\text{ }Q=-2500p+b\hfill & \text{Substitute in the point }Q=84,000\text{ and }p=30\hfill \\ 84,000=-2500\left(30\right)+b\hfill & \text{Solve for }b\hfill \\ \text{ }b=159,000\hfill & \hfill \end{array}[/latex]

This gives us the linear equation [latex]Q=-2,500p+159,000[/latex] relating cost and subscribers. We now return to our revenue equation.

[latex]\begin{array}{c}\text{Revenue}=pQ\hfill \\ \text{Revenue}=p\left(-2,500p+159,000\right)\hfill \\ \text{Revenue}=-2,500{p}^{2}+159,000p\hfill \end{array}[/latex]

We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.

[latex]\large \begin{array}{c}h=-\frac{159,000}{2\left(-2,500\right)}\hfill \\ \text{ }=31.8\hfill \end{array}[/latex]

The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.

[latex]\begin{array}{c}\text{maximum revenue}&=&-2,500{\left(31.8\right)}^{2}+159,000\left(31.8\right)\hfill \\ \text{ }&=&2,528,100\hfill \end{array}[/latex]

Analysis of the Solution

This could also be solved by graphing the quadratic. We can see the maximum revenue on a graph of the quadratic function.