Chapter 1.3, Problem 69PS

Solution

To find: length of rope needed to enclose the new section.

The new dimension of rope for enclose the new section is $\text{30 foot long by 27 foot wide}\text{.}$

Given:

A rectangular section of land measures $18\text{foot}$ by $15\text{foot}$ .

Concept Used:

Factoring method to find the roots of the quadratic equation.

Calculation:

Length of rectangle $=18\text{foot}$ .

Width of rectangle $=15\text{foot}$ .

  $\text{Area of rectangle}=\text{length }\times \text{ width}$ ,

  $\begin{array}{c}\text{Area of rectangle}=18\times 15\\ =270{\text{foot}}^2\end{array}$

Now, we have to triple the area:

New area of rectangle is $270\times 3{\text{foot}}^2=810{\text{ foot}}^2$ .

After increasing the same distance of $x\text{feet}$ to the length and the width:

New length of rectangle is $\left(x+18\right)\text{foot}$ .

New width of rectangle is $\left(x+15\right)\text{foot}$ .

Here, Applying the area of rectangle formula,

  $\therefore \left(x+18\right)\left(x+15\right)=810$

Solving this equation:

  $\begin{array}{c}\left(x+18\right)\left(x+15\right)=810\\ x\left(x+15\right)+18\left(x+15\right)=810\\ x^2+15x+18x+270=810\\ x^2+33x+270-810=0\\ x^2+33x-540=0\end{array}$

Applying the factoring method of quadratic equation:

  $\begin{array}{c}{x}^2+45x-12x-540=0\\ x^2+45x-12x-540=0\\ x\left(x+45\right)-12\left(x+45\right)=0\\ \left(x+45\right)\left(x-12\right)=0\end{array}$

Using Zero Product property:

  $\begin{array}{c}x-12=0\text{or}x+45=0\\ x=12\text{or}x=-45\end{array}$

Here, Neglecting the value of $x=-45$ as it is known that distance cannot be negative.

Therefore, The value of $x\text{ is 12}\text{.}$

Dimension of rope is:

  $\begin{array}{c}\text{length is }x+18=12+18\\ =30\text{ foot}\\ \text{width is }x+15=12+15\\ =27\text{ foot}\end{array}$

Conclusion:

The new dimension of rope for enclose the new section is $\text{30 foot long by 27 foot wide}\text{.}$