Example 1
Completion requirements
| Example 1 |
Lance is planning on drawing a large portrait. He would like the sides of his paper to have a \(9 \rm{:}16\) ratio and the area to be at least \(360 \thinspace \rm{in}^2\). Determine the interval of paper lengths that Lance can use.
Let \(l\) be the length. To maintain a \(9 \rm{:}16\) ratio, the width will need to be \(\frac{9}{16}l\), and the area will be \(l \left( \frac{9}{16} \right)\), so \(\frac{9}{{16}}l^2 \ge 360\).
Select a method to solve the inequality. The zeros and test points method is shown.
In all of the earlier solutions, the inequality was rearranged to make one side zero. Why is that an unnecessary step in this solution?
The intervals that may be part of the solution are:
The test point \(–30\) satisfies the original inequality, so \(l \le - 8 \sqrt {10}\) is part of the solution set. However, because this interval only includes negative values, it is excluded from the solution to Lance’s problem, which should only include positive values for length.
Test \(-8\sqrt {10} \le l \le 8 \sqrt{10}\). The test point \(0\) is shown.
The test point \(0\) does not satisfy the original inequality, so \(-8\sqrt{10} \le l \le 8 \sqrt{10}\) is not part of the solution set.
Test \(l \ge 8\sqrt {10}\). The test point \(30\) is shown.
The test point \(30\) satisfies the original inequality, so \(l \ge 8\sqrt {10}\) is part of the solution set.
The set {\(l \thinspace | \thinspace l \ge 8\sqrt{10}, \thinspace l \in \rm{R}\)} represents all the lengths Lance can use.
Select a method to solve the inequality. The zeros and test points method is shown.
\[\begin{align}
\frac{9}{{16}}l^2 &= 360 \\
l^2 &= 640 \\
l &= \pm \sqrt {640} \\
l &= \pm 8\sqrt {10}
\end{align}\]
\frac{9}{{16}}l^2 &= 360 \\
l^2 &= 640 \\
l &= \pm \sqrt {640} \\
l &= \pm 8\sqrt {10}
\end{align}\]
In all of the earlier solutions, the inequality was rearranged to make one side zero. Why is that an unnecessary step in this solution?
The intervals that may be part of the solution are:
- \(l \le - 8\sqrt {10}\)
- \(- 8\sqrt {10} \le l \le 8 \sqrt{10}\)
-
\(l \ge 8\sqrt {10}\)
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
\frac{9}{{16}}l^2 \\ \frac{9}{{16}}\left( {- 30} \right)^2 \\ 506.25 \end{array}\] |
\(360\)
|
|
LS \(\gt\) RS |
|
The test point \(–30\) satisfies the original inequality, so \(l \le - 8 \sqrt {10}\) is part of the solution set. However, because this interval only includes negative values, it is excluded from the solution to Lance’s problem, which should only include positive values for length.
Test \(-8\sqrt {10} \le l \le 8 \sqrt{10}\). The test point \(0\) is shown.
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
\frac{9}{{16}}l^2 \\ \frac{9}{{16}}\left( {0} \right)^2 \\ 0 \end{array}\] |
\(360\)
|
|
LS \(\lt\) RS |
|
The test point \(0\) does not satisfy the original inequality, so \(-8\sqrt{10} \le l \le 8 \sqrt{10}\) is not part of the solution set.
Test \(l \ge 8\sqrt {10}\). The test point \(30\) is shown.
| Left Side | Right Side |
|---|---|
|
\[\begin{array}{r}
\frac{9}{{16}}l^2 \\ \frac{9}{{16}}\left( {30} \right)^2 \\ 506.25 \end{array}\] |
\(360\)
|
|
LS \(\gt\) RS |
|
The test point \(30\) satisfies the original inequality, so \(l \ge 8\sqrt {10}\) is part of the solution set.
The set {\(l \thinspace | \thinspace l \ge 8\sqrt{10}, \thinspace l \in \rm{R}\)} represents all the lengths Lance can use.