Chapter 9.2, Problem 27PPE

To determine

The width that gives the maximum area, the maximum area and range of the function.

Answer to Problem 27PPE

  $\text{The maximum point }\left(Vertex\right)\text{is }\left(25,625\right).$

Maximum area = 625 ft.

Range of the function: $0\le A\le 625$

Explanation of Solution

Given:

You have 100 ft of string to rope off a rectangular section for a bake sale at school fair.

Function that gives the area of the section in square feet, where x is width in feet is:

  $A=-x^2+50x$

Calculation:

Compare the given function with $y=a{x}^2+bx+c$ .

  $\begin{array}{l}here;\\ a=-1\\ b=50\\ c=0\end{array}$

As the coefficient of $x^2$ is negative, the parabola will open downwards. The vertex will be the maximum point.

  $\begin{array}{l}\text{Equation of axis of symmetry is:}\\ x=-\frac{b}{2a}\end{array}$

Therefore, the axis of symmetry is:

  $\begin{array}{l}x=-\frac{b}{2a}\\ x=-\frac{\left(50\right)}{2\left(-1\right)}\\ x=25\end{array}$

The x-coordinate of the given function is $x=25$ . Now, the A-coordinate will be:

  $\begin{array}{l}A=-x^2+50x\\ A=-{\left(25\right)}^2+50\left(25\right)\\ A=625\\ So,\text{ the maximum point }\left(Vertex\right)\text{is }\left(25,625\right).\end{array}$

The rope will reach maximum area i.e. 625 feet when the width is 25 ft.

The graph will be:

  

The range of the function will be:

  $0\le A\le 625$