Chapter 2.2, Problem 66E

To determine

To find: The width of the football field and the maximum height of the surface of the field.

Answer to Problem 66E

The width of the football field is 160 feet.

The maximum height of the surface of the field is 1.4976 feet.

Explanation of Solution

Given Information: The cross section of a field is modelled by $y=-0.000234x\left(x-160\right)$ where x and y are measured in feet.

Calculation:

To find the width of the football field first find the x-intercepts and then find the difference between them to find the distance. Substitute $y=0$ .

  $\begin{array}{c}y=-0.000234x\left(x-160\right)\\ 0=-0.000234x\left(x-160\right)\\ x=0,160\end{array}$

The distance between the lowest and the highest x-intercept is,

  $\begin{array}{c}D=160-0\\ =160\end{array}$

• Here, D is the width of the football field.

The maximum height of the surface of the field is given by the y-coordinate of the vertex.

The x -coordinate of the vertex of a parabola of the equation $y=a{x}^2+bx+c$ is ,

  $x=-\frac{b}{2a}$

The equation $y=-0.000234x\left(x-160\right)$ in the form $y=a{x}^2+bx+c$ is,

  $\begin{array}{c}y=-0.000234x\left(x-160\right)\\ =-0.000234x^2+0.03744\end{array}$

On comparing,

  $\begin{array}{l}a=-0.000234\\ b=0.03744\\ c=0\end{array}$

The x -coordinate is,

  $\begin{array}{c}x=-\frac{0.03744}{2\left(-0.000234\right)}\\ =\frac{37440}{2\times 234}\\ =\frac{37440}{468}\\ =80\end{array}$

The maximum height therefore is,

  $\begin{array}{c}y=-0.000234x\left(x-160\right)\\ =-0.000234\times 80\times \left(80-160\right)\\ =0.000234\times 6400\\ =1.4976\end{array}$