Chapter 1.2, Problem 53PS

Solution

a.

To find: The width of football field.

The width of football field is $160$ ft.

Given data: The equation of the quadratic function representing the football field is an inverted parabola given by $y\left(x\right)=-0.000234x\left(x-160\right)$, x in feet, shown here:

  

Method/Formula used:

The intercepts made by the quadratic function ( inverted parabola) are obtained by solving the equation $y\left(x\right)=0$ . The axis of symmetry divides the parabola into two equal parts.

Calculation:

The given quadratic function or the curve representing boundary of the football field whose graph is shown in Fig. (1), is

  $y\left(x\right)=-0.000234x\left(x-160\right)$

The intercepts made by the curve on the x-axis are:

  $\begin{array}{l}-0.000234\left(x-0\right)\left(x-160\right)=0\\ \Rightarrow \left(x-0\right)\left(x-160\right)=0\\ \Rightarrow x-0=0\\ \Rightarrow x=0\\ \Rightarrow x-160=0\\ \Rightarrow x=160\end{array}$

That shoes boundary curve intersects x -axis at points $\left(0,0\right),\left(160,0\right)$ . Therefore $\begin{array}{l}\left(w,0\right)=\left(160,0\right)\\ \Rightarrow w=160\end{array}$

From the Fig. (1), w is the width of the football field.

Thus, the width of football field is $160$ ft.

b.

To find: The maximum height of the field’s surface.

The height of the football field is $1.50$ ft.

Given:

The information in part (a) Method used:

The axis of symmetry of the graph of a quadratic function is given by $x=\frac{p+q}{2}$ , where $\left(p,l\right)$ and $\left(q,l\right)$ are the intercepts of function on the x -axis.

If $x=h$ , is the axis of symmetry, then coordinates of vertex V are $\left(h,H\right)$ where $H=y\left(h\right)$ . Here, H is the height of the football field.

Calculations:

As calculated in part (a), the intercepts of the function $y\left(x\right)$ on the x -axis are $\left(0,\text{0}\right)$ and $\left(160,\text{0}\right)$ , therefore, $p=0,q=160$ .

The equation of axis of symmetry is

  $\begin{array}{c}x=\frac{0+160}{2}\\ =80\text{ft}\text{.}\end{array}$

Now, substitute $x=80$ in the equation $y\left(x\right)=-0.000234x\left(x-160\right)$ ,

  $\begin{array}{c}y\left(80\right)=-0.000234\left(80\right)\left(80-160\right)\\ =-0.000234\left(80\right)\left(-80\right)\\ =1.4976\\ \approx 1.50\end{array}$

But $y\left(80\right)$ equals H , the height of the field.

Therefore, the height of the football field is $1.50$ ft.