Graph the system of inequalities. Name the coordinates of the vertices of the feasible region

feasible-region

#1

Graph the system of inequalities. Name the coordinates of the vertices of the feasible region. Find the maximum and minimum values of the given function 5 ≥ y ≥ ­-3,
4x + y ≤ 5,
-­2x + y ≤ 5,
f(x, y) = 4x-3y

Answer:

First find the intersections.
From
4x + y <= 5
-2x + y <= 5

Rearrange:
y <= 5 - 4x
y <= 5 + 2x
Find the intersections. 5 - 4x = 5 + 2x
-4x = 2x
x = 0,
y = 5 - 40 = 5
Intersect at (0, 5)
Find the regions where they intersect Choose test points such as x = 1
y <= 5 - 4
1 = 1
y <= 5 + 21 = 7
Test x = 2
y <= 5 - 4
2 = -3
y < 5 + 2*2 = 9

It appears that the points (1, 1) and (2, -3) satisfy both inequalities. therefore we see the set of points in the interval (0, +infinity) can work.

All points below the line y = 5 - 4x on (0, +infinity)
and all points below the line y = 5 + 2x on (-infinity, 0)

basically from x =0 to x=2
and from x = -4 to x = 0
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The interval region for this solution is x = -4 to x = 2 and y = -3 to y = 5
Our corners of the feasibility region are (0,5) , (-4, -3), and (2, -3)

f(0,5) = 40 - 35 = -15
f(-4,-3) = 4*(-4) - 3*(-3) = -16 + 9 = -7
f(2, -3) = 42 - 3(-3) = 8 + 9 = 17

Our Maximum is at (2, -3) and our Minimum is at (0,5)