Math Problem Statement

what are the maximum points for the highest value of z

Solution

Let's calculate the maximum value of z=10x+10yz = 10x + 10y by first finding the corner points of the feasible region formed by the constraints. After that, we will evaluate the value of zz at each of these points.

Constraints Recap:

  1. 8x+6y1208x + 6y \geq 120
  2. 14x11y3814x - 11y \geq 38
  3. x+y42x + y \leq 42
  4. x0x \geq 0
  5. y0y \geq 0

Step 1: Find the Intersection Points

We'll find the points of intersection for the pairs of constraint lines to determine the vertices of the feasible region.

1. Intersection of 8x+6y=1208x + 6y = 120 and 14x11y=3814x - 11y = 38:

  • Solve these two equations simultaneously.

Equation 1: 8x+6y=120\text{Equation 1: } 8x + 6y = 120 Equation 2: 14x11y=38\text{Equation 2: } 14x - 11y = 38

Multiply Equation 1 by 11 and Equation 2 by 6 to eliminate yy:

88x+66y=132088x + 66y = 1320 84x66y=22884x - 66y = 228

Add these equations:

172x=1548172x = 1548 x=9x = 9

Substitute x=9x = 9 back into Equation 1:

8(9)+6y=12072+6y=1206y=48y=88(9) + 6y = 120 \rightarrow 72 + 6y = 120 \rightarrow 6y = 48 \rightarrow y = 8

So, the intersection point is (9,8)(9, 8).

2. Intersection of 8x+6y=1208x + 6y = 120 and x+y=42x + y = 42:

  • Substitute y=42xy = 42 - x into 8x+6y=1208x + 6y = 120:

8x+6(42x)=1208x + 6(42 - x) = 120 8x+2526x=1202x=132x=668x + 252 - 6x = 120 \rightarrow 2x = -132 \rightarrow x = -66

Since x=66x = -66 is not within the feasible region x0x \geq 0, this intersection does not contribute to a valid vertex in the feasible region.

3. Intersection of 14x11y=3814x - 11y = 38 and x+y=42x + y = 42:

  • Substitute y=42xy = 42 - x into 14x11y=3814x - 11y = 38:

14x11(42x)=3814x - 11(42 - x) = 38 14x462+11x=3825x=500x=2014x - 462 + 11x = 38 \rightarrow 25x = 500 \rightarrow x = 20

Substitute x=20x = 20 into y=42xy = 42 - x:

y=4220=22y = 42 - 20 = 22

So, the intersection point is (20,22)(20, 22).

4. Intersection of 8x+6y=1208x + 6y = 120 and y=0y = 0 (substituting y=0y = 0):

  • Substitute y=0y = 0 into 8x+6y=1208x + 6y = 120:

8x+0=120x=158x + 0 = 120 \rightarrow x = 15

So, the intersection point is (15,0)(15, 0).

5. Intersection of x=0x = 0 and 8x+6y=1208x + 6y = 120:

  • Substitute x=0x = 0 into 8x+6y=1208x + 6y = 120:

6y=120y=206y = 120 \rightarrow y = 20

So, the intersection point is (0,20)(0, 20).

Step 2: Evaluate z=10x+10yz = 10x + 10y at Each Vertex

The vertices of the feasible region are:

  1. (9,8)(9, 8)
  2. (20,22)(20, 22)
  3. (15,0)(15, 0)
  4. (0,20)(0, 20)

Now, evaluate zz at each vertex:

  • z(9,8)=10(9)+10(8)=90+80=170z(9, 8) = 10(9) + 10(8) = 90 + 80 = 170
  • z(20,22)=10(20)+10(22)=200+220=420z(20, 22) = 10(20) + 10(22) = 200 + 220 = 420
  • z(15,0)=10(15)+10(0)=150+0=150z(15, 0) = 10(15) + 10(0) = 150 + 0 = 150
  • z(0,20)=10(0)+10(20)=0+200=200z(0, 20) = 10(0) + 10(20) = 0 + 200 = 200

Conclusion:

The maximum value of z=420z = 420 occurs at the point (20,22)(20, 22).

Would you like any additional details or explanations?


Follow-up Questions:

  1. Would you like to see the feasible region graphically?
  2. Are you interested in verifying these calculations for the other constraints?
  3. Do you need further clarification on how the vertices were determined?
  4. Should we explore how changes in constraints affect the maximum zz value?
  5. Would you like to learn more about different methods for solving linear programming problems?

Tip:

In linear programming, the maximum or minimum value of the objective function always occurs at one of the vertices of the feasible region. Checking each vertex systematically ensures you find the optimal solution.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Linear Programming
Feasible Region
Objective Function
Constraints

Formulas

Objective function formula: z = 10x + 10y

Theorems

-

Suitable Grade Level

Advanced High School