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.

Given the objective function is and the contraints as follows:

       Substitute 0 for in .

      So, the intersecting point is .

       Substitute 0 for in .

      So, the intersecting point is .

       Substitute 0 for in .

      So, the intersecting point is .

       Substitute 0 for in .

      So, the intersecting point is .

        Add times to .

       Substitute in :

      So, the intersecting point is .

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.

Evaluate the objective function at these boundary points:

At :

At :

At:

At:

From the above calculated values, one can notice that the maximum value of is 15 and it is obtained at .

Hence, the maximum value of is 15 occurs at the corner point .

Lear more about the maximizing problems here: link

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

i belive those are the ones that are under 90 degreese.

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w