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) = 4x3y
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  41 = 1
y <= 5 + 21 = 7
Test x = 2
y <= 5  42 = 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)