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Given the objective function C=3x−2y and constraints x≥0, y≥0, 2x+y≤10, 3x+2y≤18, identify the corner point at which the maximum value of C occurs.

Respuesta :

Answer:

Step-by-step explanation:

Find  the maximum value of

C = 3x -2y  Objective function

subject to the following constraints.

Constraints

x ≥ 0

y ≥ 0                

2x + y ≤ 10  vertex 1 : when x=0 then y=10  (0,10)

3x + 2y ≤ 18 vertex 2 : y=0, then x=6          ( 6,0)

two equations together to determine vertex 3 :

3x+2y = 18

2x+y = 10

x=2, y= 6

The feasible region determined by the constraints is

shown. The three vertices are (0, 10),  and (6, 0), (0,9)

and (2,6)

First evaluate C = 3x -2 y at each of the vertices.

At (0, 10): C = 3(0) - 2(10) = -17

At (6, 0): C = 3(6) - 2(0) = 18

At ( 2,6) : C = 3(2) -2(6) = -6

At (0,9) : C = 3(0)-2(9)= -18

the maximum value occur on 18 when x=9 and y=0

Ver imagen swan85

Given the objective function is [tex]C=3x-2y[/tex] and the contraints as follows:

[tex]x\geq 0\\ y\geq 0\\ 2x+y\leq 10\\ 3x+2y\leq 18[/tex]

  • Find the intersecting point of [tex]x=0[/tex] and [tex]2x+y=10[/tex].

       Substitute 0 for [tex]x[/tex] in [tex]2x+y=10[/tex].

        [tex]2(0)+y=10\\ y=10[/tex]

      So, the intersecting point is [tex](0,10)[/tex].

  • Find the intersecting point of [tex]y=0[/tex] and [tex]2x+y=10[/tex].

       Substitute 0 for [tex]y[/tex] in [tex]2x+y=10[/tex].

        [tex]2x+0=10\\ 2x=10\\ x=5[/tex]

      So, the intersecting point is [tex](5,0)[/tex].

  • Find the intersecting point of [tex]x=0[/tex] and [tex]3x+2y=18[/tex].

       Substitute 0 for [tex]x[/tex] in [tex]3x+2y=18[/tex].

        [tex]3(0)+2y=18\\ 2y=18\\ y=9[/tex]

      So, the intersecting point is [tex](0,9)[/tex].

  • Find the intersecting point of [tex]y=0[/tex] and [tex]3x+2y=18[/tex].

       Substitute 0 for [tex]y[/tex] in [tex]3x+2y=18[/tex].

        [tex]3x+2(0)=18\\ 3x=18\\ x=6[/tex]

      So, the intersecting point is [tex](6,0)[/tex].

  • Find the intesecting point of [tex]2x+y=10,3x+2y=18[/tex].

        Add [tex]-2[/tex] times [tex]2x+y=10[/tex] to [tex]3x+2y=18[/tex].

        [tex]-2(2x+y)+3x+2y=-2(10)+18\\ -4x-2y+3x+2y=-20+18\\ -x=-2\\ x=2[/tex]

       Substitute [tex]x=2[/tex] in [tex]2x+y=10[/tex]:

        [tex]2(2)+y=10\\ 4+y=10\\y=6[/tex]

      So, the intersecting point is [tex](2,6)[/tex].

  • The origin [tex](0,0)[/tex] is also a intersecting point of [tex]x\geq 0,y\geq 0[/tex].

Corner points:

          The corner points are the boundary points of the bounded region of   the given constraints.

The bounded region of the given constraints is shown below.

  • From the graph notice that the shaded region is the required bounded region of the given constraints.
  • The boundary points are [tex](0,0),(5,0),(0,9),(2,6)[/tex].

Evaluate the objective function [tex]C=3x-2y[/tex]at these boundary points:

At [tex](0,0)[/tex]:

[tex]C=3(0)-2(0)\\C=0[/tex]

At [tex](5,0)[/tex]:

[tex]C=3(5)-2(0)\\C=15[/tex]

At[tex](0,9)[/tex]:

[tex]C=3(0)-2(9)\\C=-18[/tex]

At[tex](2,6)[/tex]:

[tex]C=3(2)-2(6)\\C=6-12\\C=-6[/tex]

From the above calculated values, one can notice that the maximum value of [tex]C[/tex] is 15 and it is obtained at [tex](5,0)[/tex].

Hence, the maximum value of [tex]C[/tex] is 15 occurs at the corner point [tex](5,0)[/tex].

Lear more about the maximizing problems here: https://brainly.com/question/13112754

Ver imagen throwdolbeau

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