Maharashtra Board Class 11 Maths Part 2 Chapter 8 Linear Inequations Miscellaneous Solutions

Linear Inequations Class 11 Commerce Maths 2 Chapter 8 Miscellaneous Exercise 8 Answers Maharashtra Board

Std 11 Maths 2 Miscellaneous Exercise 8 Solutions Commerce Maths

Solve The Following System Of Inequalities Graphically.

Question 1. \(x \geq 3\), \(y \geq 2\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(x \geq 3\)	\(x = 3\)	-	-	\(0 \geq 3\) \(\therefore\) R.H.S. of line \(x = 3\)
\(y \geq 2\)	\(y = 2\)	-	-	\(0 \geq 2\) \(\therefore\) above line \(y = 2\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो असमानताओं \(x \geq 3\) और \(y \geq 2\) के ग्राफिकल समाधान को दर्शाता है। इसमें एक निर्देशांक प्रणाली पर दो रेखाएँ \(x=3\) (एक ऊर्ध्वाधर रेखा) और \(y=2\) (एक क्षैतिज रेखा) खींची गई हैं। वह क्षेत्र जो दोनों असमानताओं को संतुष्ट करता है, यानी \(x=3\) के दाईं ओर और \(y=2\) के ऊपर का क्षेत्र, छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: To solve these inequalities graphically, draw the lines \(x=3\) and \(y=2\) on a coordinate plane. The region where \(x\) values are 3 or greater and \(y\) values are 2 or greater forms the feasible region, which is shaded.

🎯 Exam Tip: Ensure precise drawing of boundary lines and accurate shading of the feasible region, as both are crucial for scoring.

Question 2. \(3x + 2y \leq 12\), \(x \geq 1\), \(y \geq 2\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(3x + 2y \leq 12\)	\(3x + 2y = 12\)	\(\frac{x}{4} + \frac{y}{6} = 1\)	A (4, 0), B (0, 6)	\(3(0) + 2(0) \leq 12\) \(\therefore 0 \leq 12\) \(\therefore\) origin side
\(x \geq 1\)	\(x = 1\)	-	-	\(0 \geq 1\) \(\therefore\) R.H.S. of line \(x = 1\)
\(y \geq 2\)	\(y = 2\)	-	-	\(0 \geq 2\) \(\therefore\) above line \(y = 2\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख तीन असमानताओं \(3x + 2y \leq 12\), \(x \geq 1\), और \(y \geq 2\) के ग्राफिकल समाधान को दर्शाता है। इसमें तीन रेखाएँ \(3x + 2y = 12\), \(x=1\), और \(y=2\) खींची गई हैं। वह क्षेत्र जो इन तीनों असमानताओं को एक साथ संतुष्ट करता है, यानी तीनों रेखाओं द्वारा निर्धारित शर्तों के भीतर का उभयनिष्ठ क्षेत्र, छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the boundary lines for \(3x + 2y = 12\), \(x=1\), and \(y=2\). The solution is the shaded region that satisfies all three conditions: below \(3x + 2y = 12\), to the right of \(x=1\), and above \(y=2\).

🎯 Exam Tip: Clearly label intersection points and the feasible region to demonstrate a complete understanding of the solution set.

Question 3. \(2x + y \geq 6\), \(3x + 4y < 12\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(2x + y \geq 6\)	\(2x + y = 6\)	\(\frac{x}{3} + \frac{y}{6} = 1\)	A (3, 0), B (0, 6)	\(2(0) + 0 \geq 6\) \(\therefore 0 \geq 6\) \(\therefore\) non-origin side
\(3x + 4y \leq 12\)	\(3x + 4y = 12\)	\(\frac{x}{4} + \frac{y}{3} = 1\)	C (4, 0), D (0, 3)	\(3(0) + 4(0) \leq 12\) \(\therefore 0 \leq 12\) \(\therefore\) origin side

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो असमानताओं \(2x + y \geq 6\) और \(3x + 4y < 12\) के ग्राफिकल समाधान को दर्शाता है। इसमें दो रेखाएँ \(2x + y = 6\) और \(3x + 4y = 12\) खींची गई हैं। वह क्षेत्र जो \(2x + y = 6\) के गैर-मूल बिंदु की ओर और \(3x + 4y = 12\) के मूल बिंदु की ओर है, उसे उभयनिष्ठ छायांकित क्षेत्र के रूप में दिखाया गया है। The shaded portion represents the graphical solution.
In simple words: Plot the lines \(2x + y = 6\) and \(3x + 4y = 12\). The solution is the region that is above or on the line \(2x + y = 6\) (away from the origin) and strictly below the line \(3x + 4y = 12\) (towards the origin).

🎯 Exam Tip: Pay attention to strict inequalities (e.g., \(<\) or \(>\)), which indicate that the boundary line itself is not included in the solution and should be drawn as a dashed line.

Question 4. \(x + y \geq 4\), \(2x - y \leq 0\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(x + y \geq 4\)	\(x + y = 4\)	\(\frac{x}{4} + \frac{y}{4} = 1\)	A (4, 0), B (0, 4)	\(0 + 0 \geq 4\) \(\therefore 0 \geq 4\) \(\therefore\) non-origin side
\(2x - y \leq 0\)	\(2x - y = 0\)	-	O (0, 0), D (2, 4)	\(2(0) - 0 \leq 0\) \(\therefore 0 \leq 0\) \(\therefore\) origin side

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो असमानताओं \(x + y \geq 4\) और \(2x - y \leq 0\) के ग्राफिकल समाधान को दर्शाता है। इसमें दो रेखाएँ \(x + y = 4\) और \(2x - y = 0\) खींची गई हैं। वह क्षेत्र जो \(x + y = 4\) के गैर-मूल बिंदु की ओर और \(2x - y = 0\) के मूल बिंदु की ओर है, उसे उभयनिष्ठ छायांकित क्षेत्र के रूप में दिखाया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(x + y = 4\) and \(2x - y = 0\). The solution area is where the region is above or on \(x + y = 4\) (away from the origin) and below or on \(2x - y = 0\) (towards the origin, passing through the origin).

🎯 Exam Tip: For inequalities passing through the origin, choose a test point not on the line (e.g., (1,0) or (0,1)) to determine the correct shading region.

Question 5. \(2x - y \geq 1\), \(x - 2y \leq -1\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(2x - y \geq 1\)	\(2x - y = 1\)	\(\frac{2x}{1} - \frac{y}{1} = 1\) i.e., \(\frac{x}{1/2} + \frac{y}{-1} = 1\)	A \((1/2, 0)\), B (0, -1)	\(2(0) - 0 \geq 1\) \(\therefore 0 \geq 1\) \(\therefore\) non-origin side
\(x - 2y \leq -1\)	\(x - 2y = -1\)	\(\frac{x}{-1} - \frac{2y}{-1} = 1\) i.e., \(\frac{x}{-1} + \frac{y}{1/2} = 1\)	C (-1, 0), D \((0, 1/2)\)	\(0 - 2(0) \leq -1\) \(\therefore 0 \leq -1\) \(\therefore\) non-origin side

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो असमानताओं \(2x - y \geq 1\) और \(x - 2y \leq -1\) के ग्राफिकल समाधान को दर्शाता है। इसमें दो रेखाएँ \(2x - y = 1\) और \(x - 2y = -1\) खींची गई हैं। वह क्षेत्र जो दोनों रेखाओं के गैर-मूल बिंदु की ओर स्थित है और दोनों असमानताओं को संतुष्ट करता है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(2x - y = 1\) and \(x - 2y = -1\). The solution is the region that satisfies both conditions: away from the origin for \(2x - y \geq 1\) and away from the origin for \(x - 2y \leq -1\).

🎯 Exam Tip: Pay close attention when rearranging equations to the double-intercept form, especially with negative constants or coefficients, to correctly identify axis intercepts.

Question 6. \(x + y \leq 6\), \(x + y \geq 4\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(x + y \leq 6\)	\(x + y = 6\)	\(\frac{x}{6} + \frac{y}{6} = 1\)	A (6, 0), B (0, 6)	\(0 + 0 \leq 6\) \(\therefore 0 \leq 6\) \(\therefore\) origin side
\(x + y \geq 4\)	\(x + y = 4\)	\(\frac{x}{4} + \frac{y}{4} = 1\)	C (4, 0), D (0, 4)	\(0 + 0 \geq 4\) \(\therefore 0 \geq 4\) \(\therefore\) non-origin side

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो असमानताओं \(x + y \leq 6\) और \(x + y \geq 4\) के ग्राफिकल समाधान को दर्शाता है। इसमें दो समानांतर रेखाएँ \(x + y = 6\) और \(x + y = 4\) खींची गई हैं। वह क्षेत्र जो इन दोनों रेखाओं के बीच में स्थित है, उसे छायांकित किया गया है, जो दोनों असमानताओं को संतुष्ट करता है। The shaded portion represents the graphical solution.
In simple words: Graph the parallel lines \(x + y = 6\) and \(x + y = 4\). The solution region is the band between these two lines, where points satisfy being below or on \(x + y = 6\) and above or on \(x + y = 4\).

🎯 Exam Tip: When dealing with parallel lines, the feasible region often forms a band or strip between them. Clearly indicate which side of each line is included in the solution.

Question 7. \(2x + y \geq 8\), \(x + 2y \geq 10\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(2x + y \geq 8\)	\(2x + y = 8\)	\(\frac{x}{4} + \frac{y}{8} = 1\)	A (4, 0), B (0, 8)	\(2(0) + 0 \geq 8\) \(\therefore 0 \geq 8\) \(\therefore\) non-origin side
\(x + 2y \geq 10\)	\(x + 2y = 10\)	\(\frac{x}{10} + \frac{y}{5} = 1\)	C (10, 0), D (0, 5)	\(0 + 2(0) \geq 10\) \(\therefore 0 \geq 10\) \(\therefore\) non-origin side

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख दो असमानताओं \(2x + y \geq 8\) और \(x + 2y \geq 10\) के ग्राफिकल समाधान को दर्शाता है। इसमें दो रेखाएँ \(2x + y = 8\) और \(x + 2y = 10\) खींची गई हैं। वह क्षेत्र जो दोनों रेखाओं के गैर-मूल बिंदु की ओर स्थित है और दोनों असमानताओं को संतुष्ट करता है, उसे छायांकित किया गया है, जो उभयनिष्ठ व्यवहार्य क्षेत्र को दर्शाता है। The shaded portion represents the graphical solution.
In simple words: Draw the lines \(2x + y = 8\) and \(x + 2y = 10\). The solution is the shaded region where points are above or on both lines, specifically the area away from the origin for both inequalities.

🎯 Exam Tip: When multiple inequalities require shading away from the origin, the feasible region is often unbounded. Ensure the boundaries and the direction of shading are clearly indicated.

Question 8. \(x + y \leq 9\), \(y > x\), \(x \geq 0\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(x + y \leq 9\)	\(x + y = 9\)	\(\frac{x}{9} + \frac{y}{9} = 1\)	A (9, 0), B (0, 9)	\(0 + 0 \leq 9\) \(\therefore 0 \leq 9\) \(\therefore\) origin side
\(y \geq x\)	\(y = x\)	-	O (0, 0), C (1, 1)	\(0 \geq 0\) \(\therefore\) origin side
\(x \geq 0\)	\(x = 0\)	-	-	R.H.S of Y-axis

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख तीन असमानताओं \(x + y \leq 9\), \(y > x\), और \(x \geq 0\) के ग्राफिकल समाधान को दर्शाता है। इसमें तीन रेखाएँ \(x + y = 9\), \(y = x\), और \(x = 0\) (Y-अक्ष) खींची गई हैं। वह क्षेत्र जो \(x + y = 9\) के मूल बिंदु की ओर, \(y = x\) के ऊपर (मूल बिंदु की ओर), और Y-अक्ष के दाईं ओर है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(x + y = 9\), \(y = x\), and \(x = 0\). The solution is the shaded region that lies below or on \(x + y = 9\), above \(y = x\) (the line \(y=x\) itself is usually dashed for strict inequality but here shown solid), and to the right of the Y-axis.

🎯 Exam Tip: Remember that \(y > x\) means the region strictly above the line \(y = x\). If the inequality is strict, the boundary line should be drawn as a dashed line.

Question 9. \(5x + 4y \leq 20\), \(x \geq 1\), \(y \geq 2\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(5x + 4y \leq 20\)	\(5x + 4y = 20\)	\(\frac{x}{4} + \frac{y}{5} = 1\)	A (4, 0), B (0, 5)	\(5(0) + 4(0) \leq 20\) \(\therefore 0 \leq 20\) \(\therefore\) origin side
\(x \geq 1\)	\(x = 1\)	-	-	\(0 \geq 1\) \(\therefore\) R.H.S. of line \(x = 1\)
\(y \geq 2\)	\(y = 2\)	-	-	\(0 \geq 2\) \(\therefore\) above the line \(y = 2\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख तीन असमानताओं \(5x + 4y \leq 20\), \(x \geq 1\), और \(y \geq 2\) के ग्राफिकल समाधान को दर्शाता है। इसमें तीन रेखाएँ \(5x + 4y = 20\), \(x=1\), और \(y=2\) खींची गई हैं। वह क्षेत्र जो \(5x + 4y = 20\) के मूल बिंदु की ओर, \(x=1\) के दाईं ओर, और \(y=2\) के ऊपर है, उसे छायांकित किया गया है, जो उभयनिष्ठ व्यवहार्य क्षेत्र को दर्शाता है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(5x + 4y = 20\), \(x=1\), and \(y=2\). The solution is the shaded region that satisfies being below or on \(5x + 4y = 20\), to the right of or on \(x=1\), and above or on \(y=2\).

🎯 Exam Tip: When multiple linear inequalities define a region, the feasible region is a convex polygon. Ensure all vertices of this polygon are correctly identified if asked.

Question 10. \(3x + 4y \leq 60\), \(x + 3y \leq 30\), \(x \geq 0\), \(y \geq 0\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(3x + 4y \leq 60\)	\(3x + 4y = 60\)	\(\frac{x}{20} + \frac{y}{15} = 1\)	A (20, 0), B (0, 15)	\(5(0) + 4(0) \leq 60\) \(\therefore 0 \leq 60\) \(\therefore\) origin side
\(x + 3y \leq 30\)	\(x + 3y = 30\)	\(\frac{x}{30} + \frac{y}{10} = 1\)	C (30, 0), D (0, 10)	\(0 + 3(0) \leq 30\) \(\therefore 0 \leq 30\) \(\therefore\) origin side
\(x \geq 0\)	\(x = 0\)	-	-	R.H.S. of Y-axis
\(y \geq 0\)	\(y = 0\)	-	-	Above X-axis

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख चार असमानताओं \(3x + 4y \leq 60\), \(x + 3y \leq 30\), \(x \geq 0\), और \(y \geq 0\) के ग्राफिकल समाधान को दर्शाता है। इसमें चार रेखाएँ \(3x + 4y = 60\), \(x + 3y = 30\), Y-अक्ष (\(x=0\)), और X-अक्ष (\(y=0\)) खींची गई हैं। वह क्षेत्र जो सभी असमानताओं को संतुष्ट करता है, यानी मूल बिंदु की ओर और पहले चतुर्थांश में स्थित है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(3x + 4y = 60\), \(x + 3y = 30\), and the axes \(x=0, y=0\). The solution is the shaded region that is towards the origin for both main inequalities and lies in the first quadrant, representing the common feasible region.

🎯 Exam Tip: The conditions \(x \geq 0\) and \(y \geq 0\) always confine the feasible region to the first quadrant of the coordinate plane.

Question 11. \(2x + y \geq 4\), \(x + y \leq 3\), \(2x - 3y \leq 6\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(2x + y \geq 4\)	\(2x + y = 4\)	\(\frac{x}{2} + \frac{y}{4} = 1\)	A (2, 0), B (0, 4)	\(2(0) + 0 \geq 4\) \(\therefore 0 \geq 4\) \(\therefore\) non-origin side
\(x + y \leq 3\)	\(x + y = 3\)	\(\frac{x}{3} + \frac{y}{3} = 1\)	C (3, 0), D (0, 3)	\(0 + 0 \leq 3\) \(\therefore 0 \leq 3\) \(\therefore\) origin side
\(2x - 3y \leq 6\)	\(2x - 3y = 6\)	\(\frac{x}{3} - \frac{3y}{6} = 1\) i.e., \(\frac{x}{3} + \frac{y}{-2} = 1\)	C (3, 0), E (0, -2)	\(2(0) - 3(0) \leq 6\) \(\therefore 0 \leq 6\) \(\therefore\) origin side

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख तीन असमानताओं \(2x + y \geq 4\), \(x + y \leq 3\), और \(2x - 3y \leq 6\) के ग्राफिकल समाधान को दर्शाता है। इसमें तीन रेखाएँ \(2x + y = 4\), \(x + y = 3\), और \(2x - 3y = 6\) खींची गई हैं। वह क्षेत्र जो तीनों असमानताओं को संतुष्ट करता है, यानी \(2x + y = 4\) के गैर-मूल बिंदु की ओर, और \(x + y = 3\) और \(2x - 3y = 6\) दोनों के मूल बिंदु की ओर है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(2x + y = 4\), \(x + y = 3\), and \(2x - 3y = 6\). The feasible region is the area that is above or on \(2x + y = 4\), below or on \(x + y = 3\), and below or on \(2x - 3y = 6\).

🎯 Exam Tip: When identifying the feasible region, carefully test a point (like the origin, if not on a boundary) for each inequality to determine the correct side for shading.

Question 12. \(x - 2y \leq 3\), \(3x + 4y \geq 12\), \(x \geq 0\), \(y \geq 1\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(x - 2y \leq 3\)	\(x - 2y = 3\)	\(\frac{x}{3} - \frac{2y}{3} = 1\) i.e., \(\frac{x}{3} + \frac{y}{-3/2} = 1\)	A (3, 0), B \((0, -3/2)\)	\(0 - 2(0) \leq 3\) \(\therefore 0 \leq 3\) \(\therefore\) origin side
\(3x + 4y \geq 12\)	\(3x + 4y = 12\)	\(\frac{x}{4} + \frac{y}{3} = 1\)	C (4, 0), D (0, 3)	\(3(0) + 4(0) \geq 12\) \(\therefore 0 \geq 12\) \(\therefore\) non-origin side
\(x \geq 0\)	\(x = 0\)	-	-	R.H.S. of Y-axis
\(y \geq 1\)	\(y = 1\)	-	-	Above the line \(y = 1\)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख चार असमानताओं \(x - 2y \leq 3\), \(3x + 4y \geq 12\), \(x \geq 0\), और \(y \geq 1\) के ग्राफिकल समाधान को दर्शाता है। इसमें चार रेखाएँ \(x - 2y = 3\), \(3x + 4y = 12\), Y-अक्ष (\(x=0\)), और \(y=1\) खींची गई हैं। वह क्षेत्र जो \(x - 2y = 3\) के मूल बिंदु की ओर, \(3x + 4y = 12\) के गैर-मूल बिंदु की ओर, Y-अक्ष के दाईं ओर, और \(y=1\) के ऊपर है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(x - 2y = 3\), \(3x + 4y = 12\), \(x=0\), and \(y=1\). The solution region is confined by these lines: towards the origin for the first, away from the origin for the second, in the first quadrant, and above the line \(y=1\).

🎯 Exam Tip: Carefully handle inequalities that define horizontal or vertical lines (like \(x \geq 0\) or \(y \geq 1\)), as they significantly constrain the feasible region to a specific quadrant or strip.

Question 13. \(4x + 3y \leq 60\), \(y \geq 2x\), \(x \geq 3\), \(x, y \geq 0\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(4x + 3y \leq 60\)	\(4x + 3y = 60\)	\(\frac{x}{15} + \frac{y}{20} = 1\)	A (15, 0), B (0, 20)	\(4(0) + 3(0) \leq 60\) \(\therefore 0 \leq 60\) \(\therefore\) origin side
\(y \geq 2x\)	\(y = 2x\)	-	O (0, 0), C (10, 20)	\(0 \geq 2(0)\) \(\therefore 0 \geq 0\) \(\therefore\) origin side
\(x \geq 3\)	\(x = 3\)	-	-	\(0 \geq 3\) \(\therefore\) R.H.S. of the line \(x = 3\)
\(x \geq 0\)	\(x = 0\)	-	-	R.H.S. of Y-axis
\(y \geq 0\)	\(y = 0\)	-	-	Above X-axis

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख पाँच असमानताओं \(4x + 3y \leq 60\), \(y \geq 2x\), \(x \geq 3\), \(x \geq 0\), और \(y \geq 0\) के ग्राफिकल समाधान को दर्शाता है। इसमें पाँच रेखाएँ \(4x + 3y = 60\), \(y = 2x\), \(x = 3\), Y-अक्ष (\(x=0\)), और X-अक्ष (\(y=0\)) खींची गई हैं। वह क्षेत्र जो सभी असमानताओं को संतुष्ट करता है, यानी \(4x + 3y = 60\) के मूल बिंदु की ओर, \(y = 2x\) के ऊपर, \(x = 3\) के दाईं ओर, और पहले चतुर्थांश में है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the lines \(4x + 3y = 60\), \(y = 2x\), \(x = 3\), and the axes. The solution is the shaded region that lies towards the origin for the first, above for the second, to the right for the third, and within the first quadrant, satisfying all conditions.

🎯 Exam Tip: The combination of \(x \geq 0, y \geq 0\) with other constraints often creates a bounded feasible region in the first quadrant. Identifying all corner points is crucial for optimization problems.

Question 14. \(3x + 2y \leq 150\), \(x + 4y \geq 80\), \(x \leq 15\), \(y \geq 0\), \(x \geq 0\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(3x + 2y \leq 150\)	\(3x + 2y = 150\)	\(\frac{x}{50} + \frac{y}{75} = 1\)	A (50, 0), B (0, 75)	\(3(0) + 2(0) \leq 150\) \(\therefore 0 \leq 150\) \(\therefore\) origin side
\(x + 4y \geq 80\)	\(x + 4y = 80\)	\(\frac{x}{80} + \frac{y}{20} = 1\)	C (80, 0), D (0, 20)	\(0 + 4(0) \geq 80\) \(\therefore 0 \geq 80\) \(\therefore\) non-origin side
\(x \leq 15\)	\(x = 15\)	-	-	\(\therefore\) L.H.S. of the line \(x = 15\)
\(x \geq 0\)	\(x = 0\)	-	-	R.H.S. of Y-axis
\(y \geq 0\)	\(y = 0\)	-	-	Above X-axis

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख पाँच असमानताओं \(3x + 2y \leq 150\), \(x + 4y \geq 80\), \(x \leq 15\), \(y \geq 0\), और \(x \geq 0\) के ग्राफिकल समाधान को दर्शाता है। इसमें पाँच रेखाएँ \(3x + 2y = 150\), \(x + 4y = 80\), \(x = 15\), Y-अक्ष (\(x=0\)), और X-अक्ष (\(y=0\)) खींची गई हैं। वह क्षेत्र जो \(3x + 2y = 150\) के मूल बिंदु की ओर, \(x + 4y = 80\) के गैर-मूल बिंदु की ओर, \(x = 15\) के बाईं ओर, और पहले चतुर्थांश में है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the boundary lines for all given inequalities: \(3x + 2y = 150\), \(x + 4y = 80\), \(x = 15\), \(x=0\), and \(y=0\). The shaded feasible region is bounded by these lines, satisfying all the specified conditions simultaneously.

🎯 Exam Tip: When \(x\) is bounded by two vertical lines (e.g., \(x \geq 0\) and \(x \leq 15\)), the feasible region will be confined within that vertical strip.

Question 15. \(x + 2y \leq 10\), \(x + y \geq 1\), \(x - y \leq 0\), \(x \geq 0\), \(y \geq 0\)
Answer:
To find a graphical solution, construct the table as follows:

Inequation	Equation	Double Intercept form	Points (x, y)	Region
\(x + 2y \leq 10\)	\(x + 2y = 10\)	\(\frac{x}{10} + \frac{y}{5} = 1\)	A (10, 0), B (0, 5)	\(0 + 2(0) \leq 10\) \(\therefore 0 \leq 10\) \(\therefore\) origin side
\(x + y \geq 1\)	\(x + y = 1\)	\(\frac{x}{1} + \frac{y}{1} = 1\)	C (1, 0), D (0, 1)	\(0 + 0 \geq 1\) \(\therefore 0 \geq 1\) \(\therefore\) non-origin side
\(x - y \leq 0\)	\(x - y = 0\)	-	O (0, 0), E (1, 1)	\(0 - 0 \leq 0\) \(\therefore 0 \leq 0\) \(\therefore\) origin side
\(x \geq 0\)	\(x = 0\)	-	-	R.H.S. of Y-axis
\(y \geq 0\)	\(y = 0\)	-	-	Above X-axis

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख पाँच असमानताओं \(x + 2y \leq 10\), \(x + y \geq 1\), \(x - y \leq 0\), \(x \geq 0\), और \(y \geq 0\) के ग्राफिकल समाधान को दर्शाता है। इसमें पाँच रेखाएँ \(x + 2y = 10\), \(x + y = 1\), \(x - y = 0\), Y-अक्ष (\(x=0\)), और X-अक्ष (\(y=0\)) खींची गई हैं। वह क्षेत्र जो सभी असमानताओं को संतुष्ट करता है, यानी \(x + 2y = 10\) के मूल बिंदु की ओर, \(x + y = 1\) के गैर-मूल बिंदु की ओर, \(x - y = 0\) के मूल बिंदु की ओर, और पहले चतुर्थांश में है, उसे छायांकित किया गया है। The shaded portion represents the graphical solution.
In simple words: Graph the boundary lines \(x + 2y = 10\), \(x + y = 1\), \(x - y = 0\), and the axes. The shaded region represents the solution that is below or on \(x + 2y = 10\), above or on \(x + y = 1\), to the left of or on \(x - y = 0\), and confined to the first quadrant.

🎯 Exam Tip: Carefully determine the intersection points of the boundary lines, as these points define the vertices of the feasible region, which are critical for optimization problems.

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