Maharashtra Board Class 12 Maths Part 2 Chapter 6 Linear Programming 6.2 Solutions

Solve the following LPP by graphical method.

Question 1. Maximize z = 11x + 8y, Subject to x ≤ 4 ,y ≤ 6 x + y ≤ 6, x ≥ 0, y ≥ 0.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें x≤4, y≤6 और x+y≤6 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र AOBE है। मूल बिंदु O(0,0), A(4,0), B(0,6) और E(4,2) सुसंगत क्षेत्र के शीर्ष बिंदु हैं, जिनका उपयोग उद्देश्य फलन z को अधिकतम करने के लिए किया जाता है।

The feasible solution is AOBE
Where A(4, 0) O(0, 0) B(0, 6)
E is the point of intersection of x + y = 6 and x = 4.
\( \therefore \) 4 + y = 6
\( \implies \) y = 2
\( \therefore \) E = (4, 2)

\( \therefore \) z is maximum at (4, 2) and the maximum value of z = 60
In simple words: This problem finds the highest possible value of 'z' (an objective function) by evaluating it at the corners of a feasible region defined by given inequalities on a graph. The point (4, 2) gave the maximum value of 60.

🎯 Exam Tip: Always plot the feasible region accurately and identify all corner points. Evaluate the objective function at each corner point to find the maximum or minimum value. Showing calculation steps for each corner point's 'z' value is crucial for full marks.

Question 2. Maximize z = 4x + 6y, Subject to 3x + 2y ≤ 12, x + y ≥ 4 x, y ≥ 0.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें 3x+2y≤12, x+y≥4, x≥0, y≥0 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र ABC है। शीर्ष बिंदु A(0,6), B(4,0), और C(0,4) हैं, जिनका उपयोग उद्देश्य फलन z को अधिकतम करने के लिए किया जाता है।

From figure, ABC is the feasible region
Where A(0, 6) B(4, 0) C(0, 4)

Maximum value of z = 36 at A(0, 6)
In simple words: This problem aims to find the maximum value of 'z' within the region defined by the given linear inequalities. By evaluating 'z' at each corner point of the feasible region, A(0, 6) yields the highest value of 36.

🎯 Exam Tip: Clearly label the feasible region on your graph. When dealing with "non-origin side" regions, ensure you correctly identify the unbounded or specific bounded area for evaluation. Accurate point identification is key for correct 'z' value calculations.

Question 3. Maximize z = 7x + 11y, Subject to 3x + 5y ≤ 26, 5x + 3y ≤ 30, x ≥ 0, y ≥ 0.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें 3x+5y≤26 और 5x+3y≤30 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र AODE है। शीर्ष बिंदु A(0, 5.2), O(0,0), D(6,0) और E(4.5, 2.5) हैं, जिनका उपयोग उद्देश्य फलन z को अधिकतम करने के लिए किया जाता है।

\( \therefore \) AODE is the feasible region where
A(0, 5.2) O(0, 0) D(6, 0) and E is the intersection of 3x + 5y = 26 and 5x + 3y = 30
For E,
Solving 3x + 5y = 26 ......(i)
5x + 3y = 30 ......(ii)
We get, x = 4.5, y = 2.5
\( \therefore \) E = (4.5, 2.5)

\( \therefore \) Maximum value of z = 59 at E(4.5, 2.5)
In simple words: This problem aims to find the maximum value of 'z' within the feasible region defined by the given constraints. After identifying all corner points, including the intersection point E, the objective function 'z' is maximized at E(4.5, 2.5) with a value of 59.

🎯 Exam Tip: Pay careful attention to solving simultaneous equations for intersection points, as these are crucial corner points for evaluating the objective function. Fractional or decimal coordinates are common in LPPs, so calculate accurately.

Question 4. Maximize z = 10x + 25y, Subject to 0 ≤ x ≤ 3, 0≤ y ≤ 3, x + y ≤ 5.
Answer:
The constraints can be written as, x ≤ 3, x ≥ 0, y ≥ 0, y ≤ 3, x + y ≤ 5

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें x≤3, y≤3 और x+y≤5 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र ABCDE है। शीर्ष बिंदु A(3,0), B(0,0), C(0,3), D(2,3) और E(3,2) हैं, जिनका उपयोग उद्देश्य फलन z को अधिकतम करने के लिए किया जाता है।

ABCDE is the feasible region where A(3, 0) B(0, 0) and C(0, 3) D is the intersection
of y = 3 and x + y = 5 and E is the intersection of x = 3 and x + y = 5
For D,
Solving y = 3 .........(i)
x + y = 5 .........(ii)
We get x = 2, y = 3
\( \therefore \) D = (2, 3)
For E,
Solving x = 3 .......(i)
x + y = 5 .........(ii)
We get x = 3, y = 2
\( \therefore \) E = (3, 2)

\( \therefore \) Maximum value of z = 90 at D(2, 3)
In simple words: This problem aims to maximize 'z' by finding the optimal point within the feasible region defined by the given linear inequalities. The maximum value of 'z' is found to be 90 at the corner point D(2, 3).

🎯 Exam Tip: When multiple boundaries define the feasible region, ensure you find all intersection points that form the vertices of this region. A thorough evaluation of 'z' at each vertex is essential to correctly identify the maximum or minimum.

Question 5. Maximize z = 3x + 5y, Subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें x+4y≤24, 3x+y≤21 और x+y≤9 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र OAGHD है। शीर्ष बिंदु O(0,0), A(0,6), G(4,5), H(6,3) और D(7,0) हैं, जिनका उपयोग उद्देश्य फलन z को अधिकतम करने के लिए किया जाता है।

OAGHD is the feasible region where O(0, 0), A(0, 6), D(7, 0) G is the intersecting
point of x + 4y = 24 and x + y = 9
H is the intersecting points of 3x + y = 21 and x + y = 9.
For G, Solving x + 4y = 24 .......(i)
x + y = 9 .........(ii)
We get, x = 4, y = 5
\( \therefore \) G (4, 5)
For H, Solving x + y = 9 .........(i)
3x + y = 21 ........(ii)
We get x = 6, y = 3
\( \therefore \) H(6, 3)

\( \therefore \) Maximum value of z = 37 at the point G(4, 5)
In simple words: This problem finds the maximum value of 'z' within a complex feasible region defined by several linear inequalities. By evaluating 'z' at each corner point, including the intersection points G and H, the highest value of 37 is obtained at G(4, 5).

🎯 Exam Tip: For problems with multiple constraints, accurately identifying all vertices of the feasible region is crucial. Use algebraic methods to find precise intersection points of boundary lines, as visual estimation can lead to errors.

Question 6. Minimize z = 7x + y Subject to 5x + y ≥ 5, x + y ≥ 3, x ≥ 0, y ≥ 0
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें 5x+y≥5, x+y≥3, x≥0, y≥0 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र AED है। शीर्ष बिंदु A(0,5), E(1/2, 5/2) और D(3,0) हैं, जिनका उपयोग उद्देश्य फलन z को न्यूनतम करने के लिए किया जाता है।

AED is the feasible region where A(0, 5) D(3, 0) and E is the point of intersection of
5x + y = 5 and x + y = 3.
For E, Solving 5x + y = 5 .........(i)
x + y = 3 ........(ii)
We get, x = \( \frac{1}{2} \), y = \( \frac{5}{2} \)
\( \therefore \) E(\( \frac{1}{2} \), \( \frac{5}{2} \))

\( \therefore \) Minimum value of z = 5 at A(0, 5)
In simple words: This problem seeks to find the lowest possible value of 'z' within the feasible region defined by the given "greater than or equal to" inequalities. By evaluating 'z' at the corner points of this unbounded feasible region, the minimum value of 5 is found at A(0, 5).

🎯 Exam Tip: For minimization problems with "greater than or equal to" constraints, the feasible region is often unbounded. Ensure you correctly identify all corner points, including any intersection points, and test them to find the true minimum value.

Question 7. Minimize z = 8x + 10y, Subject to 2x + y ≥ 7, 2x + 3y ≥ 15 ,y ≥ 2, x ≥ 0, y ≥ 0
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें 2x+y≥7, 2x+3y≥15, y≥2, x≥0, y≥0 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र AEG है। शीर्ष बिंदु A(0,7), F(3/2, 4) और G(4.5, 2) हैं, जिनका उपयोग उद्देश्य फलन z को न्यूनतम करने के लिए किया जाता है।

AEG is the feasible solution where A(0, 7)
F is the point of intersection of 2x + y = 7 and 2x + 3y = 15
G is the point of intersection of y = 2 and 2x + 3y = 15
For F, Solving 2x + y = 7 ........(i)
2x + 3y = 15 .........(ii)
We get x = \( \frac{3}{2} \), y = 4
\( \therefore \) F = (\( \frac{3}{2} \), 4)
For G, Solving 2x + 3y = 15 ........(i)
y = 2 .......(ii)
We get x = 4.5, y = 2
\( \therefore \) G = (4.5, 2)

\( \therefore \) Minimum value of z = 52 at F(\( \frac{3}{2} \), 4)
In simple words: This problem aims to find the minimum value of 'z' within the feasible region formed by multiple "greater than or equal to" constraints. The lowest value of 'z' is found to be 52 at the intersection point F(3/2, 4) of two boundary lines.

🎯 Exam Tip: When given `y ≥ constant` as a constraint, it defines a horizontal line. Be sure to consider this line and its intersections with other boundaries when determining the feasible region and its corner points for minimization.

Question 8. Minimize z = 6x + 2y, Subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आलेखीय विधि से एक रैखिक प्रोग्रामिंग समस्या का हल दर्शाता है। इसमें x+2y≥3, x+4y≥4 और 3x+y≥3 जैसी असमानताओं द्वारा परिभाषित एक सुसंगत क्षेत्र DGHB है। शीर्ष बिंदु D(0,3), G(3/5, 6/5), H(2, 1/2) और B(4,0) हैं, जिनका उपयोग उद्देश्य फलन z को न्यूनतम करने के लिए किया जाता है।

DGHB is the feasible region where D(0, 3), B(4, 0)
G is the point of intersection of 3x + y = 3 and x + 2y = 3 and H is the point of
intersection of x + 2y = 3 and x + 4y = 4
For G, Solving 3x + y = 3 ......(i)
x + 2y = 3 .........(ii)
We get x = \( \frac{3}{5} \), y = \( \frac{6}{5} \)
\( \therefore \) G(\( \frac{3}{5} \), \( \frac{6}{5} \))
For H, Solving x + 2y = 3 ........(i)
x + 4y = 4 .........(ii)
We get, x = 2, y = \( \frac{1}{2} \)
\( \therefore \) H(2, \( \frac{1}{2} \))

\( \therefore \) Minimum value of z = 22.5 at H(2, \( \frac{1}{2} \))
In simple words: This problem focuses on minimizing 'z' within an unbounded feasible region defined by multiple "greater than or equal to" inequalities. By calculating 'z' at each critical corner point, including the intersections G and H, the lowest value of 22.5 is found at H(2, 1/2).

🎯 Exam Tip: When the feasible region is complex and involves multiple intersection points, systematically solving each pair of equations to find the exact coordinates of the vertices is paramount. Double-check your calculations, especially with fractions, to avoid errors in determining the minimum or maximum value.