Chapter 9.1, Problem 49PPE

Solution

(a)

To graph: The function $y=135-x^2$ .

The required domain is 135 and the range is $\left(-11.7,11.7\right)$ .

Given Information:

The given functions is $y=135-x^2$ .

Here $x$ is the side length in feet.

  $y$ gives the area.

Calculation:

Consider the given functions $y=135-x^2$ .

Now, draw the graph.

  

Interpretation:

Hence, from the graph the domain is 135 and the range is $\left(-11.7,11.7\right)$ .

(b)

To find: The domain of the function.

The required domain is 135.

The given functions is $y=135-x^2$ .

Here $x$ is the side length in feet.

  $y$ gives the area.

Explanation:

Consider the given functions $y=135-x^2$ .

From the solution 49a. the domain of the function is the point where the graph cuts at y axis.

Thus, $y=135$ .

Hence, the required domain is 135.

(c)

To find: The rang of the function.

The required range is $\left(-11.7,11.7\right)$ .

The given functions is $y=135-x^2$ .

Here $x$ is the side length in feet.

  $y$ gives the area.

Explanation:

Consider the given functions $y=135-x^2$ .

From the solution 49a. the range of the function are the points where the graph cuts at x axis.

Thus, $\left(-11.7,11.7\right)$ .

Hence, the required range is $\left(-11.7,11.7\right)$ .

(d)

To calculate: The side lengths of the window.

The required side length are 13.96 ft by 8.38 ft.

The given previous side length of the window is 15 ft by 9 ft.

New area is $117{\text{ft}}^2$

Calculation:

Consider the new are $117{\text{ft}}^2$ .

Let the sides of the new window is $a\text{ft}\text{by}b\text{ft}$ .

Thus, the area is,

  $ab=117{\text{ft}}^2$  ...... (1)

The side scale of the new window must be same as the previous one.

Now, the scale of the previous sides $\frac{9}{15}=\frac{3}{5}$

The scale of the new sides $\frac{b}{a}$ .

Thus, $\frac{b}{a}=\frac{3}{5}$

Rewrite the expression as,

  $\begin{array}{c}\frac{b}{a}=\frac{3}{5}\\ b=\frac{3a}{5}\end{array}$  ...... (2)

Now, solve (1) and (2) for $a\text{and}b$ .

Substitute $b=\frac{3a}{5}$ in the equation (1).

  $\begin{array}{c}a\times \frac{3a}{5}=117\\ a^2=117\times \frac{5}{3}\\ a=\sqrt{195}\\ \approx 13.96\end{array}$

Substitute $a=13.96$ in the equation (2).

  $\begin{array}{c}b=\frac{3\times 13.96}{5}\\ \approx 8.38\end{array}$

Hence, the required side length are 13.96 ft by 8.38 ft.