Question

Solve the given Linear Programming Problem Graphically.
Maximize Z = 60X + 50Y
Subjected to,
2X + 4Y ≤ 80
3X + 2Y ≤ 60
X, Y ≥ 0

Answer 1

Answer:

400

Step-by-step explanation:

Function to maximize: Z=60x+40y

Constraints: x+2y≤12

2x+y≤12

x+

4

5

y≥5

x≥0,y≥0

Cornor Points Values of Z=60x+40y at cornor points

(0,4) 160

(0,6) 240

(4, 4) 400 ← Maximum

(6,0) 360

(5,0) 300

Hence, the maximum value of Z=60x+40y is 400.

Answer 2

Step-by-step explanation:

Given:

$Z = 60x + 50y$

subjected to

$2x + 4y \leqslant 80 \\ 3x + 2y \leqslant 60$

$xy \geqslant 0$

To find: Solve the given Linear Programming Problem Graphically. To maximize Z = 60X + 50Y

Solution:

Step 1: Write equations from given in-equations.

$2x + 4y = 80...eq1$

$3x + 2y = 60...eq2$

Step 2: Draw the lines.

For eq1

(0,20) and (40,0) are y and x intercept respectively.

For eq2

(0,30) and (20,0) are y and x intercept respectively.

Draw lines.As the in-equations holds origin,so region below the lines and origin(x-axis and y- axis) are included.

Step 3: Write the points of optimization region.

These are

(0,0)

(0,20)

(10,15)

(20,0)

*See the attached graph.

Step 4: Find the value of Z for all the points.

$\bf \red{Z_{(0,0)} = 0}$

$Z_{(0,20)} = 60 \times 0 + 50 \times 20$

$\bf \red{Z_{(0,20)} = 1000}$

$Z_{(10,15)}= 60 \times 10 + 50 \times 15$

$Z_{(10,15)} = 600 + 750$

$\bf \green{Z_{(10,15)} = 1350}$

$Z_{(20,0)}= 60 \times 20 + 50 \times 0$

$\bf \red{Z_{(20,0)} = 1200}$

Therefore,

The function Z is maximum at x=10 and y =15,and maximum value is 1350.

Final answer:

The function Z is maximum at x=10 and y =15,and maximum value is 1350.

Hope it will help you.

Learn more:

1) Solve the following Linear Programming Problem graphically: Minimise Z = - 3x + 4 y subject to x + 2y ≤ 8, 3x + 2y ≤ 12,...

https://brainly.in/question/8208599

2) Solve the following Linear Programming Problem graphically: Maximise Z = 3x + 4y subject to the constraints : x + y ≤ 4,...

https://brainly.in/question/8208597