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# find the coordinates of the ordered pair where the max. value occurs for the equal. p=2x+5y given constraints of:

x+2yless than or equal to 8

3x+yless than or equal to 9

x greater than or equal to 0

y greater than or equal to 0

### 2 Answers by Expert Tutors

Nataliya D. | Patient and effective tutor for your most difficult subject.Patient and effective tutor for your mos...
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So, we have system of four inequalities
x + 2y ≤ 8
3x + y ≤ 9
y ≥ 0
x ≥ 0
First we have to solve first two for "y"
y ≤ (-1/2)x + 4
y ≤ - 3x + 9
Second, draw the solid lines of two functions
y = (-1/2)x + 4
y = -3x + 9 and shadow the area under those two lines.
Inequalities x ≥ 0 and y ≥ 0 indicate the first quadrant
Finally, the solution of our system (all marked areas overlapped) is the tetragon (quadrilateral) with coordinates of the vertexes
(0,0) , (0, 4) , (2,3) and (3,0)
If we plug in all those coordinates, one-by-one, into p = 2x + 5y , we will see that pmax will be at (0,4)

Gene G. | You can do it! I'll show you how.You can do it! I'll show you how.
5.0 5.0 (257 lesson ratings) (257)
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This problem is more riddle than math.  You will need to graph three equations to visualize the situation:

Constraints:
(1) x + 2y <= 8  ==> plot y <= -0.5x + 4 (area below the line is allowed)
(2) 3x + y <= 9  ==> plot y <= -3x + 9

p = 2x + 5y:  (Use p = 25 to visualize the problem.)
(3) plot y = -0.4x + 5

First look at the slopes of the lines:
Line (1) has m = -0.5
Line (2) has m = -3
x >= 0 and y >= 0 restrict us the quadrant I.

Line (3) has m = -0.4 (slightly less negative than line (1)), and b=p/5.

When you plot these you can see that the line (3) for p=25 is completely above line (1) in quadrant I, and its slope is less negative than the constraint line, so it will never go below it.  We can shift line (3) up or down by changing its y-intercept.  The best we can do is to make it intersect the y axis at the same point as line (1). That intercept is y=4.  Look at the original formula for p=... and you can see that p must be 20 make the intercept for line (3) be 4.

Replace line (3) with y = -0.4x + 4, which corresponds to p = 20.
Plot this line.  It crosses the y-axis at the same point as line (1), and is thereafter always above it. Since line (3) moves down as we decrease the value of p further, 20 is the highest allowable value for p, and (0,4) is where this occurs.