Math Problem Statement

Maximize p = 2x + 3y subject to 3x + 8y ≤ 48, 6x + 4y ≤ 42, x ≥ 0, y ≥ 0.

Solution

The problem shown is a linear programming (LP) problem, where you need to maximize the objective function p=2x+3yp = 2x + 3y subject to the given constraints:

3x+8y483x + 8y \leq 48 6x+4y426x + 4y \leq 42 x0,y0x \geq 0, \quad y \geq 0

Steps to solve:

  1. Graph the constraints:

    • First, convert each inequality into an equation to determine the boundary lines.

    • For 3x+8y=483x + 8y = 48, the intercepts are:

      • x=0x = 0, y=6y = 6 (solve for yy when x=0x = 0)
      • y=0y = 0, x=16x = 16 (solve for xx when y=0y = 0)
    • For 6x+4y=426x + 4y = 42, the intercepts are:

      • x=0x = 0, y=10.5y = 10.5 (solve for yy when x=0x = 0)
      • y=0y = 0, x=7x = 7 (solve for xx when y=0y = 0)
  2. Find the feasible region:

    • Plot the lines and shade the region that satisfies all constraints, including x0x \geq 0 and y0y \geq 0.
    • The feasible region will be a polygon where these lines intersect.
  3. Find the corner points of the feasible region:

    • Solve the system of equations at the intersection points to determine the corner points of the polygon.
    • The intersection of 3x+8y=483x + 8y = 48 and 6x+4y=426x + 4y = 42 can be solved simultaneously.
  4. Evaluate the objective function at each corner point:

    • Calculate p=2x+3yp = 2x + 3y for each of the vertices of the feasible region.
    • The maximum value of pp at these points will be the solution to the maximization problem.

I'll go ahead and calculate the corner points and the corresponding values of pp.

Solving for intersection points:

  • Intersection of 3x+8y=483x + 8y = 48 and 6x+4y=426x + 4y = 42:

    Let's solve these equations using substitution or elimination.

From 6x+4y=426x + 4y = 42, divide by 2: 3x+2y=21(Equation 1)3x + 2y = 21 \quad \text{(Equation 1)}

Now, subtract Equation 1 from 3x+8y=483x + 8y = 48: (3x+8y)(3x+2y)=4821(3x + 8y) - (3x + 2y) = 48 - 21 6y=27y=276=4.56y = 27 \quad \Rightarrow \quad y = \frac{27}{6} = 4.5

Substitute y=4.5y = 4.5 into Equation 1: 3x+2(4.5)=213x+9=213x=12x=43x + 2(4.5) = 21 \quad \Rightarrow \quad 3x + 9 = 21 \quad \Rightarrow \quad 3x = 12 \quad \Rightarrow \quad x = 4

So, the intersection point is (x,y)=(4,4.5)(x, y) = (4, 4.5).

Corner points:

The corner points of the feasible region are:

  1. (0,6)(0, 6)
  2. (4,4.5)(4, 4.5)
  3. (7,0)(7, 0)
  4. (0,0)(0, 0)

Evaluate p=2x+3yp = 2x + 3y at each corner point:

  1. At (0,6)(0, 6): p=2(0)+3(6)=18p = 2(0) + 3(6) = 18
  2. At (4,4.5)(4, 4.5): p=2(4)+3(4.5)=8+13.5=21.5p = 2(4) + 3(4.5) = 8 + 13.5 = 21.5
  3. At (7,0)(7, 0): p=2(7)+3(0)=14p = 2(7) + 3(0) = 14
  4. At (0,0)(0, 0): p=0p = 0

Conclusion:

The maximum value of pp is 21.5, which occurs at the point (4,4.5)(4, 4.5).

Would you like further details on this, or do you have any questions?

Here are 5 related questions:

  1. How do we handle linear programming problems with more than two variables?
  2. Can you explain the graphical method in more detail?
  3. What happens if the feasible region is unbounded?
  4. How do we identify if a problem has multiple optimal solutions?
  5. Can we use software tools to solve larger linear programming problems?

Tip: In linear programming, always check the boundary and intersection points of the constraints to maximize or minimize the objective function efficiently.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Graphical Method
Optimization

Formulas

p = 2x + 3y
3x + 8y ≤ 48
6x + 4y ≤ 42

Theorems

Feasible Region in Linear Programming
Corner Point Theorem

Suitable Grade Level

Grades 10-12, University Intro