Chapter 2.1, Problem 60E

a.

To determine

To write:The total area A of play areas of child-care centers as a function of ‘x’ .

Answer to Problem 60E

The total area of the child-care center as a function of ‘x’ is $A\left(x\right)=2x\left(100-2x\right)$ sq. feet.

Explanation of Solution

Given:

The view of child care center is shown in Figure-1. Total fencing of the center is 200 feet.

  

Formula/ concept used:

Area of rectangle is A= length $A=\text{length×width}$

Calculations:

The total fencing of child-care center $2\left(x+x\right)+2\left(y\right)=200$ feet or $2x+y=100$ feet.

Total area of the center

  $A\left(x\right)=\text{2}\left(x\cdot y\right)=2x\left(100-2x\right)$$\because y=\text{100-2}x$

Thus, total area of the child-care center as a function of ‘x’ is $A\left(x\right)=2x\left(100-2x\right)$ sq. feet.

Conclusion:

The total area of the child-care center as a function of ‘x’ is $A\left(x\right)=2x\left(100-2x\right)$ sq. feet.

b.

To determine

To create:A table showing possible values of “x ” and the corresponding total area A of the plays areas, and estimate dimensions that will produce the maximum enclosed area.

Answer to Problem 60E

The total enclosed area of child-care center is maximum $A_{\max .}=2500$ sq. ft. when $x=y=25$ ft. and $y\text{=50}$ ft.

Explanation of Solution

Given:

The expression for total area of the play areas as function of x : $A\left(x\right)=2x\left(100-2x\right)$ .

Concept used:

We choose values of x arbitrarily such that A is positive.

Calculations:

The expression for Ais$A\left(x\right)=200x-4x^2$ .

The table for total area of the play areas A and x is given below:

x (in feet)	5	10	15	20	25	30	35
A(in sq. ft)	900	1600	2100	2400	2500	2400	2100

From the table, we observe that total area A is maximum $A_{\max .}=2500$ sq. ft. when $x=25$ ft.

For $x=25$ , $y=\text{100-2}x=\text{100-2×25=50}$ ft.

Thus, the total enclosed area of child-care center is maximum $A_{\max .}=2500$ sq. ft. when $x=y=25$ ft. and $y\text{=50}$ ft.

Conclusion:

The total enclosed area of child-care center is maximum $A_{\max .}=2500$ sq. ft. when $x=25$ ft. and $y\text{=50}$ ft.

c.

To determine

To approximate:The dimensions of child-care center so that total play area is maximum, using graphing utility.

Answer to Problem 60E

Using graphing utility, the total area $A\left(x\right)$ is maximum equal to 2500 sq. ft. when $x=25$ ft. and $y==50$ ft.

Explanation of Solution

Given:

The expression $A\left(x\right)=200x-4x^2$ in part (b)

Method used:

We use graphing calculator.

Graph:

The graph of A versus x is shown in Figure-2 here.

  

We see that the total area $A\left(x\right)$ is maximum equal to 2500 sq. ft. when $x=25$ ft.

  $\Rightarrow y=100-2x=100-2\cdot 25=50$ ft.

Conclusion:

Using graphing utility, the total area $A\left(x\right)$ is maximum equal to 2500 sq. ft. when $x=25$ ft. and $y==50$ ft.

d.

To determine

To write:The area $A\left(x\right)$ function in standard form to find algebraically dimensions that will produce the maximum enclosed area.

Answer to Problem 60E

The total area $A\left(x\right)$ is maximum equal to 2500 sq. ft. when $x=25$ ft. and $y==50$ ft.

Explanation of Solution

Given:

The area function $A\left(x\right)=200x-4x^2$ .

Formula used:

The standard form of quadratic function that represent a parabola is $f\left(x\right)=a{\left(x-h\right)}^2+k$ with vertex $\left(h,k\right)$ . If “a ” is negative the parabola opens downwards and maxima occurs at the vertex, therefore, $f_{\max .}=k$

Calculations:

The areafunction is $A\left(x\right)=200x-4x^2$ that can be written as:

  $A\left(x\right)=-4\left(x^2-50x\right)$

  $\Rightarrow A\left(x\right)=-4\left(x^2-2\cdot x\cdot 25+625\right)+2500$

  $\Rightarrow A\left(x\right)=-4{\left(x-25\right)}^2+2500$

Thus, $A\left(x\right)$ represents a parabola with vertex $\left(25,2500\right)$ .

Since, $a=-4<0$ so $A_{\max .}=2500$ sq. ft. when $x=25\Rightarrow y=100-2x=100-2\cdot 25=50$ ft.

Conclusion:

The total area $A\left(x\right)$ is maximum equal to 2500 sq. ft. when $x=25$ ft. and $y==50$ ft.

e.

To determine

To compare: The results from parts (b), (c), and (d).

Answer to Problem 60E

The results obtained parts (b), (c), and (d) are identical.

Explanation of Solution

Given:

The results obtained in parts (b), (c), and (d).

Explanations:

In parts (b), (c), and (d) we find that the area $A_{\max .}=2500$ sq. ft. when $x=25$ and $y=50$ ft.

Thus, the results obtained parts (b), (c), and (d) are identical.