Chapter 1.9, Problem 8MRPS

Solution

To determine the value of $x$ given in the problem and to explain why the other value of $x$ is rejected.

The value of $x$ is found:

  $x=\frac{-39+\sqrt{1041}}{-6}$ .

The, other value of $x$ is rejected since it exceeds both the dimensions of the paper.

Given:

Given the diagram of the paper and its stripes:

  

Given that the area of the stripes is one third of the area of the paper.

Given the dimensions of the paper:

  $5\text{in }\times \text{8}\text{in}$ .

Concept Used:

Area of a rectangle:

Given the dimension of a rectangle: $a\text{unit}\times \text{ b}\text{unit}$ .

Then, the area of the rectangle:

  $ab{\text{unit}}^2$ .

Quadratic formula:

Given a quadratic equation $a{x}^2+bx+c=0$ .

Then, the solutions of the equation:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Find the area of the paper:

Given the dimensions of the paper:

  $5\text{in }\times \text{8}\text{in}$

Then, the area of the paper $A_p$ :

  $\begin{array}{l}{A}_p=5\times 8{\text{in}}^2\\ A_p=40{\text{in}}^2\end{array}$ .

Find the area of the stripes:

It is given that the area of the stripes $A_s$ is one third of that of the paper.

That is:

  $\begin{array}{l}{A}_s=\frac{1}{3}\times A_p{\text{in}}^2\\ A_s=\frac{1}{3}\cdot 40{\text{in}}^2\\ A_s=\frac{40}{3}{\text{in}}^2\end{array}$ .

That is, the area of the strips:

  $A_s=\frac{40}{3}{\text{in}}^2$ .

Find an algebraic expression for the area of the stripes:

  

The given diagram is redrawn.

The area of the vertical stripe:

  $8x{\text{in}}^2$ .

The area of the horizontal stripe:

  $5x{\text{in}}^2$ .

Now, the area of the stripes, $A_s$ :

  $\begin{array}{l}{A}_s=\text{Sum of the areas of the individual stripes -The area of the intersecting portion}\\ A_s=\left(8x+5x\right){\text{in}}^2\text{ -}\left(x^2\right){\text{in}}^2\\ A_s=13x-x^2{\text{in}}^2\end{array}$ .

Thus, the algebraic form of the area of the stripes:

  $A_s=13x-x^2{\text{in}}^2$

Find the value of $x$by solving $A_s=\frac{40}{3}{\text{in}}^2$and $A_s=13x-x^2{\text{in}}^2$:

Substitute $A_s=\frac{40}{3}{\text{in}}^2$ in $A_s=13x-x^2{\text{in}}^2$ :

  $\begin{array}{l}\frac{40}{3}{\text{in}}^2=13x-x^2{\text{in}}^2\\ \frac{40}{3}=13x-x^2\\ 40=3\left(13x-x^2\right)\\ 40=39x-3x^2\\ 0=-3x^2+39x-40\end{array}$ .

Thus, it is found that:

  $-3x^2+39x-40=0$ .

Then, the solutions of the equation:

  $\begin{array}{l}x=\frac{-39\pm \sqrt{{\left(-39\right)}^2-4\cdot \left(-3\right)\cdot \left(-40\right)}}{2\cdot \left(-3\right)}\\ x=\frac{-39\pm \sqrt{1041}}{-6}\\ x=\frac{-39\pm \sqrt{1041}}{-6}\end{array}$ .

Thus:

  $x=\frac{-39+\sqrt{1041}}{-6}$ and $x=\frac{-39-\sqrt{1041}}{-6}$ .

Now, $\frac{-39-\sqrt{1041}}{-6}\approx 11.87$ .

But, the value of $x$ cannot exceed any dimension of the paper.

That is, $x=\frac{-39-\sqrt{1041}}{-6}$ is not the appropriate solution.

Then, it follows that, $x=\frac{-39+\sqrt{1041}}{-6}$ .

Conclusion:

The value of $x$ is found:

  $x=\frac{-39+\sqrt{1041}}{-6}$ .

The, other value of $x$ is rejected since it exceeds either dimensions of the paper.