SOLUTION: Maximize z=4x+5y subject to x+y\le 4 x-y\ge -2 x\le 3 x\ge 0 y\ge 0

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Question 1202520: Maximize z=4x+5y
subject to
x+y\le 4
x-y\ge -2
x\le 3
x\ge 0
y\ge 0
Found 2 solutions by math_tutor2020, Edwin McCravy:
Answer by math_tutor2020(3817)   (Show Source): You can put this solution on YOUR website!

Plot the five inequalities shown using graphing software such as Desmos or GeoGebra.
The overlapped shaded region is a quadrilateral with these five corner points
(0,0)
(0,2)
(1,3)
(3,1)
(3,0)

Test each corner point with the objective function to see which gives the largest value of z.

Testing (0,0)
z=4x+5y
z=4*0+5*0
z=0

Now try (0,2)
z=4x+5y
z=4*0+5*2
z=10
This result is larger than the previous.
So far (0,2) yields the largest z value.

Now try (1,3)
z=4x+5y
z=4*1+5*3
z=19
This result is larger than the previous.

Now try (3,1)
z=4x+5y
z=4*3+5*1
z=17
This result is NOT larger than the previous. So we ignore this.

Lastly we need to try (3,0)
z=4x+5y
z=4*3+5*0
z=12
This result is NOT larger than 19, so we ignore this.


Answer:
The max value is z = 19.
It occurs when (x,y) = (1,3)
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!


Did you mean ? I suppose so.
Multiply it through by -1 and change
the direction of the inequality.




The last two inequalities means we are in the first quadrant.

So we draw the three boundary lines, which are just like the inequalities, 
except we replace the symbol of inequality with an " = ".  

So we draw  in red, green, blue, and purple
respectively



Pick the origin (x,y) = (0,0) x=0, y=0 as a test point in the inequality for
each of the boundary lines to determine whether the shading is above or below
a slanted or horizontal boundary line or right or left or a vertical bountary line.

Substitute x=0, y=0 in ,  get  which is true, so the
origin is a solution, and is below the line, so all solutions are on or below
the line x+y=4.  So we are to shade below the line x+y=4.

Substitute x=0, y=0 in ,  get  which is true, so the
origin is a solution, and is below the line, so all solutions are on or below
the line -x+y=2.  So we are to shade below the line x+y=4.

Substitute x=0, y=0 in ,  get  which is true, so the
origin is a solution, and is left of the line, so all solutions are on or left
of the line x=3.  So we are to shade left the line x=3.

So we are to shade the region which is below all three lines. So



There are 5 corner points in the shaded region: 
(0,0), (0,2), (1,3), (3,1), (3,0)

The maximum and minimum values of z=4x+5y will be at one of these:

Evaluating z=4x+5y at (0,0) gives z = 4(0)+5(0) = 0+0 = 0.
Evaluating z=4x+5y at (0,2) gives z = 4(0)+5(2) = 0+10 = 10
Evaluating z=4x+5y at (1,3) gives z = 4(1)+5(3) = 4+15 = 19.
Evaluating z=4x+5y at (3,1) gives z = 4(3)+5(1) = 12+5 = 17
Evaluating z=4x+5y at (3,0) gives z = 4(3)+5(0) =12+0 = 12.

So it turns out that the maximum value of z is 19 when x=1 and y=3.

Edwin
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