Chapter 11.1, Problem 35WE

(a)

To determine

To Explain: Why the other side is $(20-x)m$ long.

Answer to Problem 35WE

  $20-xm$

Explanation of Solution

Given: Suppose you have $40m$ of fencing with which to make a rectangular pen for a dog. If one side of the rectangle is $x$ m long.

  

Perimeter of the pen is $40m$

  $\begin{array}{l}P=2(l+b)\\ 40=2(l+b)\\ 20=l+b\\ l+b=20\\ x+b=20\\ b=20-xm\end{array}$

Hence, the other side is $20-xm$

(b)

To determine

To Find: The area of pen in terms of $x$ .

Answer to Problem 35WE

  $A=20x-x^2$

Explanation of Solution

Given: Suppose you have $40m$ of fencing with which to make a rectangular pen for a dog. If one side of the rectangle is $x$ m long.

  

Length= $20-xm$

Width= $x$ m

Thus

Area of the pen

  $\begin{array}{l}A=lb\\ A=(20-x)(x)\\ A=20x-x^2\end{array}$

Hence, $A=20x-x^2$

(c)

To determine

To Find: The area of pen for each value of $x=0,2,4,6,8,10,12,16,18,20$ . Record the answers in the axes.

Explanation of Solution

Given: Suppose you have $40m$ of fencing with which to make a rectangular pen for a dog. If one side of the rectangle is $x$ m long.

  

Area of the pen for $x=0,2,4,6,8,10,12,16,18,20$

  $A=20x-x^2$

Substitute the values of $x$ in the area.

The table shows the area for given $x$ values.

  

The area on the axes is shown below.

  

(d)

To determine

To Find: The dimensions of the pen with the greatest area.

Answer to Problem 35WE

  $x=10$

Explanation of Solution

Given: Suppose you have $40m$ of fencing with which to make a rectangular pen for a dog. If one side of the rectangle is $x$ m long.

  

When $x=10$

The pen has the greatest area $A=100sq.units$

Hence, when $x=10$