Math Problem Statement

Solve the linear programming problem. Maximize and minimize z equals 4 x plus 5 y Subject to 2 x plus y greater than or equals 30 x plus 2 y greater than or equals 24 x comma y greater than or equals 0 . . . Question content area right Part 1 What is the maximum value of​ z? Select the correct choice below and fill in any answer boxes present in your choice.   A. zequals    enter your response here ​(Type an integer or a​ fraction.) B. There is no maximum value of z.

Solution

Linear Programming Problem:

You want to:

  • Maximize and minimize z=4x+5yz = 4x + 5y

Subject to the constraints:

  1. 2x+y302x + y \geq 30
  2. x+2y24x + 2y \geq 24
  3. x0,y0x \geq 0, \, y \geq 0 (non-negativity constraints)

Step 1: Convert the inequalities to equalities (for graphing)

Rearrange the constraints to get:

  1. 2x+y=302x + y = 30
  2. x+2y=24x + 2y = 24

Step 2: Graph the inequalities

Plot these two lines on the xy-plane and identify the feasible region by finding the points where the inequalities are satisfied.

Step 3: Identify corner points

The solution to the linear programming problem will be at one of the vertices (corner points) of the feasible region. The corner points are obtained by:

  • Finding the intersections of the lines.
  • Using the intercepts with the x- and y-axes.

Intersection of the lines:

Solve 2x+y=302x + y = 30 and x+2y=24x + 2y = 24 simultaneously. From the first equation, solve for yy: y=302xy = 30 - 2x

Substitute this into the second equation: x+2(302x)=24x + 2(30 - 2x) = 24 x+604x=24x + 60 - 4x = 24 3x=36x=12-3x = -36 \quad \Rightarrow \quad x = 12

Now substitute x=12x = 12 back into y=302xy = 30 - 2x: y=302(12)=6y = 30 - 2(12) = 6 Thus, the intersection point is (12,6)(12, 6).

x-intercept and y-intercept of the lines:

  • For 2x+y=302x + y = 30:

    • x-intercept: Set y=0y = 0, then x=15x = 15.
    • y-intercept: Set x=0x = 0, then y=30y = 30.
  • For x+2y=24x + 2y = 24:

    • x-intercept: Set y=0y = 0, then x=24x = 24.
    • y-intercept: Set x=0x = 0, then y=12y = 12.

Step 4: Test the corner points

The corner points of the feasible region are:

  • (15,0)(15, 0)
  • (0,12)(0, 12)
  • (12,6)(12, 6)

Now, evaluate z=4x+5yz = 4x + 5y at each of these points:

  1. At (15,0)(15, 0): z=4(15)+5(0)=60z = 4(15) + 5(0) = 60

  2. At (0,12)(0, 12): z=4(0)+5(12)=60z = 4(0) + 5(12) = 60

  3. At (12,6)(12, 6): z=4(12)+5(6)=48+30=78z = 4(12) + 5(6) = 48 + 30 = 78

Step 5: Conclusion

The maximum value of zz is 7878, which occurs at the point (12,6)(12, 6).

Final Answer:

A. z=78z = 78

Would you like further explanation or details on this process?


Here are 5 related questions to expand your understanding:

  1. How would the solution change if we wanted to minimize zz?
  2. What is the geometric interpretation of the feasible region in linear programming?
  3. How do slack variables work in converting inequalities into equalities for linear programming?
  4. Can you solve this problem graphically by sketching the region of constraints?
  5. How does the Simplex Method approach linear programming compared to the graphical method?

Tip: In linear programming, the optimal solution always lies at a corner (vertex) of the feasible region.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Graphical Method
Inequalities

Formulas

Objective Function: z = 4x + 5y
Constraints: 2x + y ≥ 30, x + 2y ≥ 24
Intersection of lines for solving: 2x + y = 30 and x + 2y = 24

Theorems

Corner Point Theorem in Linear Programming

Suitable Grade Level

Grades 10-12