Linear Programming Problem: Maximize Function P10x5y Subject

Linear Programming Problemmaximize Function P10x5ysubject To

Analyze a linear programming problem that aims to maximize the function P = 10x + 5y under the constraints 3x + 6y ≤ 18, 2x – y ≤ 2, with x ≥ 0 and y ≥ 0. The problem requires graphing the boundary lines, shading feasible regions, identifying corner points, calculating the objective function at each point, and determining the point where P reaches its maximum. Additionally, address set theory operations with given sets, analyze elements in a Venn diagram, count combinations of fast food orders, construct passwords under specific constraints, simplify factorial expressions, compute permutations and combinations, and evaluate probabilities in a committee selection scenario. This comprehensive assessment covers linear programming, set theory, combinatorics, and probability, emphasizing both theoretical understanding and practical applications.

Paper For Above instruction

The problem of maximizing a linear function subject to certain constraints is a fundamental topic in linear programming, an optimization technique used extensively in operations research and management science. The specific problem posed here involves maximizing the function P = 10x + 5y while respecting the constraints 3x + 6y ≤ 18, 2x – y ≤ 2, with x and y non-negative. Addressing this problem step by step reveals the systematic approach to solving linear programming problems using graphical methods.

First, the boundary lines of the constraints must be graphed. The line 3x + 6y = 18 can be rewritten as y = (18 – 3x)/6 or y = 3 – 0.5x. This line intersects the axes at points when x=0 (y=3) and y=0 (x=6). The second constraint, 2x – y = 2, can be rearranged to y = 2x – 2. It intersects the axes at (x=0, y=–2) (which is outside the feasible region since y ≥ 0) and at (x=1, y=0). This line slopes upwards and intersects the feasible region when x ≥ 1. The non-negativity constraints x ≥ 0 and y ≥ 0 further bound the feasible region to the first quadrant.

Next, shading the feasible region involves identifying the intersection of all inequalities—areas where all conditions are simultaneously satisfied. Plotting these lines, the feasible region is bounded by the axes, the line y=3–0.5x, and the line y=2x–2, constrained to the first quadrant. Once the feasible region is shaded, the corner points or vertices are identified because, in linear programming, the maximum or minimum of the objective function occurs at vertices. These vertices are found at the intersections of the boundary lines, which include (x=0, y=0), the intersection of y=3–0.5x with y=2x–2, and the intersection of y=3–0.5x with the axes, among others calculated algebraically.

Solving for the intersection of y=3–0.5x and y=2x–2:

- Set 3 – 0.5x = 2x – 2

- 3 + 2 = 2x + 0.5x

- 5 = 2.5x

- x = 2

- Substitute back to find y: y = 3 – 0.5*2 = 3 – 1 = 2

- The intersection point is (2, 2).

Next, evaluate the objective function P at each corner point:

- At (0, 0): P = 100 + 50 = 0

- At (0, 3): P = 0 + 53 = 15, but check if feasible: 30 + 63= 18, satisfies constraint; also, 20–3=–3 ≤ 2, satisfies second constraint, and x,y ≥ 0, so feasible.

- At (2, 2): P= 102 + 52= 20 + 10= 30

- At (6, 0): P= 106 + 50= 60, but check constraints: 36 + 60= 18 ≤18, OK. 2*6– 0= 12 ≤ 2? No, 12 > 2, so this point is outside the feasible region.

Thus, the maximum P=30 occurs at point (2, 2). Therefore, the optimal solution is x=2, y=2, with maximum profit P=30.

Beyond linear programming, the problem extends to set theory with specific sets S1 and S2: {1, 5, 6, 10} and {5, 6, 10, 12}. The union S1 ∪ S2 combines all elements from both sets without duplication: {1, 5, 6, 10, 12}. The intersection S1 ∩ S2 includes common elements: {5, 6, 10}.

In the given Venn diagram with total elements 30, blue subset A has 10 elements, yellow subset B has 15, and 5 elements are common to both A and B. The number of elements in A ∪ B is calculated as:

- |A ∪ B| = |A| + |B| – |A ∩ B| = 10 + 15 – 5= 20 elements.

The number of elements in A \ B (elements in A but not in B) is:

- |A| – |A ∩ B| = 10 – 5= 5.

Similarly, the complement of the intersection in the union, (A∩B)’ within the union, includes elements in either A or B but not both, totaling:

- |A ∪ B| – |A ∩ B| = 20 – 5= 15.

The problem also discusses counting the number of ways to order food, creating passwords, simplifying factorial expressions, permutations, combinations, and computing probabilities. For example, ordering options involve multiply counts: 3 soups × 5 sandwiches × 6 drinks= 90 possible meal combinations.

Constructing passwords with two different letters followed by two different digits involves the counting rule:

- Number of choices for the first letter: 26

- For the second letter (different from the first): 25

- For the first digit: 10

- For the second digit (different from the first): 9

Total passwords: 26 × 25 × 10 × 9= 58,500.

Simplification of factorial expressions like 8!/6! can be achieved by recognizing:

- 8! / 6! = (8×7×6!) / 6! = 8×7=56.

Choosing teams involves permutations, with the number of ways to order 5 teams selecting 3 for first, second, and third place:

- Permutations: P(5,3)= 5×4×3= 60.

Selecting 4 shirts from 10, regardless of order, involves combinations:

- C(10,4)= (10×9×8×7)/(4×3×2×1)= 210.

Finally, the probability problems of selecting committee members involve combinatorial calculations. For example, selecting at least three freshmen involves summing probabilities of selecting exactly 3 or 4 freshmen from the group, calculated using combinations divided by total possibilities:

- Total possible 4-person committees: C(15,4)= 1365.

- Exactly 3 freshmen: C(4,3)×C(11,1)= 4×11=44.

- Exactly 4 freshmen: C(4,4)= 1.

- Probability at least 3 freshmen: (44+1)/1365 ≈ 0.0322.

In conclusion, this extensive set of problems encompasses linear programming, set operations, combinatorics, and probability, each requiring methodical application of mathematical principles to solve real-world and theoretical questions effectively.

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