Question

topic: writing inequalities from a real world problem. graphing inequalities. 16 on a final for a creative writing course, ben was required to write a combination of at least 10 poems or paragraphs. ben knew that each poem would take him 30 minutes to write while a paragraph would only take 10 minutes. ben was given two hours to complete the exam. a. write an inequality to model each constraint. (hint: one constraint is time and the other is the number of needed items. let x be the number of poems written and y be the number of paragraphs written.) b. graph each inequality on a separate coordinate grid and shade the solution set for each.

Explanation:

Part a: Writing Inequalities

Step 1: Define Variables and Time Conversion

Let \( x \) be the number of poems and \( y \) be the number of paragraphs. Convert 2 hours to minutes: \( 2 \times 60 = 120 \) minutes. The time for poems is \( 30x \) minutes, and for paragraphs is \( 10y \) minutes. The total time must be less than or equal to 120 minutes, so \( 30x + 10y \leq 120 \). Also, the total number of items (poems + paragraphs) must be at least 10, so \( x + y \geq 10 \). Additionally, \( x \geq 0 \) and \( y \geq 0 \) (non - negative number of items).

Step 2: Simplify the Time Inequality

Divide \( 30x + 10y \leq 120 \) by 10: \( 3x + y \leq 12 \).

Step 1: Graph \( x + y \geq 10 \)

Rewrite it as \( y \geq -x + 10 \). The boundary line \( y=-x + 10 \) has a slope of - 1 and a y - intercept of 10. Since the inequality is \( \geq \), draw a solid line. Test the point (0,0): \( 0+0=0
ot\geq10 \), so shade above the line.

Step 2: Graph \( 3x + y \leq 12 \)

Rewrite it as \( y\leq - 3x+12 \). The boundary line \( y = - 3x + 12 \) has a slope of - 3 and a y - intercept of 12. Since the inequality is \( \leq \), draw a solid line. Test the point (0,0): \( 0 + 0=0\leq12 \), so shade below the line.

Step 3: Graph \( x\geq0 \) and \( y\geq0 \)

\( x\geq0 \) is the region to the right of the y - axis (solid line \( x = 0 \), shade right). \( y\geq0 \) is the region above the x - axis (solid line \( y = 0 \), shade above).

Step 4: Find the Feasible Region

The feasible region is the intersection of all shaded regions. It is a polygon (or a region bounded by the intersection of the lines \( x + y=10 \), \( 3x + y = 12 \), \( x = 0 \), and \( y = 0 \)). To find the intersection of \( x + y=10 \) and \( 3x + y=12 \), subtract the first equation from the second: \( (3x + y)-(x + y)=12 - 10\Rightarrow2x=2\Rightarrow x = 1 \). Substitute \( x = 1 \) into \( x + y=10 \), we get \( y=9 \). So the intersection point is (1,9). The other vertices of the feasible region can be found by intersecting with the axes: for \( x + y=10 \), when \( x = 0 \), \( y = 10 \); when \( y = 0 \), \( x = 10 \) (but \( 3x+y\leq12 \) when \( x = 10 \), \( 3\times10+y\leq12\Rightarrow y\leq - 18 \), which is not in the non - negative region). For \( 3x + y=12 \), when \( x = 0 \), \( y = 12 \); when \( y = 0 \), \( x = 4 \). The feasible region is bounded by (1,9), (4,0), (0,12) (but we also have the \( x + y\geq10 \) constraint, so we need to check the overlap). The actual feasible region vertices are (1,9), (2,8), (3,9) (wait, no, let's re - check. When \( x = 1 \), \( y = 9 \); \( x = 2 \), \( 3\times2+y\leq12\Rightarrow y\leq6 \), and \( x + y\geq10\Rightarrow2 + y\geq10\Rightarrow y\geq8 \), which is a contradiction. Wait, there is a mistake in the earlier intersection. Let's solve \( x + y=10 \) and \( 3x + y=12 \) again. \( 3x + y-(x + y)=12 - 10\Rightarrow2x = 2\Rightarrow x = 1 \), \( y=9 \). Now, check \( x = 0 \): from \( x + y\geq10 \), \( y\geq10 \); from \( 3x + y\leq12 \), \( y\leq12 \). So when \( x = 0 \), \( y \) ranges from 10 to 12. When \( y = 0 \): from \( x + y\geq10 \), \( x\geq10 \); from \( 3x + y\leq12 \), \( x\leq4 \), which is a contradiction. So the feasible region is a line segment between (1,9) and (0,10) (wait, when \( x = 0 \), \( y = 10 \) satisfies \( x + y=10 \) and \( 3\times0 + 10=10\leq12 \)) and between (1,9) and (4,0) (wait, when \( x = 4 \), \( y = 0 \) satisfies \( 3\times4+0 = 12\leq12 \) and \( 4 + 0=4
ot\geq10 \), so no). Actually, the correct feasible region is where \( x + y\geq10 \), \( 3x + y\leq12 \), \( x\geq0 \), \( y\geq0 \). The only points that satisfy all are when \( x = 1 \), \( y = 9 \); \( x = 0 \), \( y = 10 \); \( x = 2 \), \( y = 8 \) (wait, \( 3\times2+8 = 14>12 \), no). Wait, I made a mistake in the time conversion. The problem says "at least 10 poems or paragraphs", so it's \( x + y\geq10 \), and time is 30 minutes per poem and 10 minutes per paragraph, total time 2 hours (120 minutes). So \( 30x + 10y\leq120 \). Let's take \( x = 0 \): \( 10y\leq120\Rightarrow y\leq12 \), and \( y\geq10 \), so \( y\in[10,12] \). \( x = 1 \): \( 30+10y\leq120\Rightarrow10y\leq90\Rightarrow y\leq9 \), and \( 1 + y\geq10\Rightarrow y\geq9 \), so \( y = 9 \). \( x =…

Answer:

(Inequalities):

• \( x + y \geq 10 \) (total items constraint)
• \( 3x + y \leq 12 \) (time constraint)
• \( x \geq 0 \)
• \( y \geq 0 \)

Part b: Graphing Inequalities