Optimization (inequality)

Question

The following model represents an optimization problem. Determine the maximum solution. 

What is the Optimization model:

 

Restrictions: x, y is an element of real numbers 

Constraints:

x≥0

y≥0
x+2y≤6
2x-y<3

Objective function: K=x-3y

Answer 1

I would recommend getting graph paper. First we need the graph of the region.

 

The first two constraints limit the region to the first quadrant, including the x and y axis.

 

The 3rd constraint produces the line inequality y <=  -1/2X + 3   when solving for y in terms of x. <-- 1st line

 

The 4th constraint produces the line inequality 2x -3 < y when solving for y in terms of x.   <---- 2nd line

 

Graphing these two lines while focusing on the first quadrant, the region is obtained by

plugging in the origin (0,0) into the two line inequalities to determine the direction of shading.

 

For the first line, the origin produces a true statement of 0 <= -1/2(0) + 3 or simply 0 <= 3.

The second line, the origin produces a true statement of 2(0)-3 < 0 or simply -3 < 0.

 

So the shading is BELOW the line -1/2X + 3 in the first quadrant (in the southwest direction)

while above the line 2x-3 in the first quadrant (in the northwest direction)

 

Finally, the point of intersection is needed along with the roots of the two line equations:

The lines meet when

  -1/2X + 3 = 2x - 3

 

Moving everything to the right side:

  0 = 5/2X - 6

 

 0 = 5x - 12  <--- multiplies everything by 2

 

 X = 12/5 --->  y = 2(12/5) - 3 = 24/5 - 3 = 24/5 - 15/5 = 9/5

 

So the intersection point is (12/5,9/5)

 

The first line has root x=6 which is outside the shaded region.

 

THe second line has root 3/2.

 

So the shaded region contains 4 vertices:

 (0,3)   <--- the y-intercept of the first line

 

(12/5,9/5)   <-- the intersection point

 

(3/2,0) <-- the root of the second line

 

(0,0)

 

 

 

To complete the optimization, we plug each of the vertices into the objective function.

 

the origin (0,0) gives a result of 0.

(0,3) gives a result of -9

 

(12/5,9/5) gives a result of 12/5 - 3(9/5) = 12/5 - 27/5 = -15/5 = -3

 

(3/2,0) gives a result of 3/2 - 3(0) = 3/2

 

So the objective function is maximized when (3/2,0) under these constraints