Math Problem Statement

Maximize P = 100x + 50y subject to these constraints: 20x + 35y ≤ 280 10x + 10y ≤ 86 20x + 15y ≤ 140 40x + 15y ≤ 196 50x + 10y ≤ 230 x ≥ 0, y ≥ 0 Maximum value for P =

Solution

To solve this linear programming problem, we will use the method of graphical analysis or apply the Simplex method. The objective is to maximize P=100x+50yP = 100x + 50y, subject to the following constraints:

  1. 20x+35y28020x + 35y \leq 280
  2. 10x+10y8610x + 10y \leq 86
  3. 20x+15y14020x + 15y \leq 140
  4. 40x+15y19640x + 15y \leq 196
  5. 50x+10y23050x + 10y \leq 230
  6. x0x \geq 0, y0y \geq 0

Steps:

  1. Convert the inequalities to equations to find corner points of the feasible region.
  2. Find the intersection points by solving pairs of equations.
  3. Evaluate the objective function P=100x+50yP = 100x + 50y at each of these corner points to find the maximum value.

Let’s solve this systematically.

Step 1: Solve pairs of equations

1. 20x+35y=28020x + 35y = 280

We can rewrite this in terms of yy: y=28020x35y = \frac{280 - 20x}{35}

2. 10x+10y=8610x + 10y = 86

Simplifying, x+y=8.6x + y = 8.6, so: y=8.6xy = 8.6 - x

3. 20x+15y=14020x + 15y = 140

Simplifying: y=14020x15y = \frac{140 - 20x}{15}

4. 40x+15y=19640x + 15y = 196

Simplifying: y=19640x15y = \frac{196 - 40x}{15}

5. 50x+10y=23050x + 10y = 230

Simplifying: y=23050x10y = \frac{230 - 50x}{10}

Step 2: Find intersection points

We now solve these equations in pairs to get the coordinates of the feasible region.

Step 3: Evaluate P=100x+50yP = 100x + 50y at these points

I'll calculate these points and evaluate PP.The intersection points (feasible region corners) and the corresponding values of P=100x+50yP = 100x + 50y are:

  1. (75,365)\left( \frac{7}{5}, \frac{36}{5} \right), P=500P = 500
  2. (74,7)\left( \frac{7}{4}, 7 \right), P=525P = 525
  3. (13355,36455)\left( \frac{133}{55}, \frac{364}{55} \right), P572.73P \approx 572.73
  4. (10531,18831)\left( \frac{105}{31}, \frac{188}{31} \right), P641.94P \approx 641.94
  5. (115,325)\left( \frac{11}{5}, \frac{32}{5} \right), P=540P = 540
  6. (6725,14825)\left( \frac{67}{25}, \frac{148}{25} \right), P=564P = 564
  7. (185,5)\left( \frac{18}{5}, 5 \right), P=610P = 610
  8. (145,285)\left( \frac{14}{5}, \frac{28}{5} \right), P=560P = 560
  9. (4111,4811)\left( \frac{41}{11}, \frac{48}{11} \right), P590.91P \approx 590.91
  10. (14935,127)\left( \frac{149}{35}, \frac{12}{7} \right), P511.43P \approx 511.43

Maximum value:

The maximum value of P641.94P \approx 641.94 occurs at the point (10531,18831)\left( \frac{105}{31}, \frac{188}{31} \right).

Would you like further details on how these points were calculated?

Here are five related questions for further exploration:

  1. How do you graph linear inequalities to find feasible regions?
  2. What is the significance of corner points in linear programming?
  3. How can the Simplex method be used to solve linear programming problems?
  4. How would changing the objective function coefficients affect the solution?
  5. What happens if the constraints change and no feasible solution exists?

Tip: In linear programming, the maximum or minimum of the objective function always occurs at one of the vertices of the feasible region.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Inequalities
Feasible Region

Formulas

Objective function: P = 100x + 50y
Constraints: 20x + 35y ≤ 280, 10x + 10y ≤ 86, 20x + 15y ≤ 140, 40x + 15y ≤ 196, 50x + 10y ≤ 230

Theorems

Corner Point Theorem
Linear Programming Feasibility Theorem

Suitable Grade Level

University Level (Linear Programming, Optimization)