Answer:
C = 14 is the minimum value
Step-by-step explanation:
Sketch the inequalities using
2x + y = 20
with x- intercept = (10, 0) and y- intercept = (0, 20)
2x + 3y = 36
with x- intercept = (18, 0) and y- intercept = (0, 12)
Solve 2x + y = 20 and 2x + 3y = 36 simultaneously to find
The point of intersection at (6, 8))
The vertices of the feasible region are at
(0, 20), (6, 8), (18, 0)
Evaluate the objective function C = x + y at each of the vertices
(0, 20) → C = 0 + 20 = 20
(6, 8) → C = 6 + 8 = 14 ← minimum value
(18, 0) → C = 18 + 0 = 18
Minimum value is C = 14 when x = 6 and y = 8
The constraints can be rewritten with the following subdivision along the x axis:
Case 1: 0 <= x <= 6
In this case, the constraint for y becomes
[tex]y \geq -2x+20[/tex]
Which implies that
[tex]x+y \geq x-2x+20=20-x[/tex]
The mimum for this function, given that x ranges from 0 to 6, is clearly 20-6=14.
Case 1: 6 <= x <= 18
In this case, the constraint for y becomes
[tex]y \geq -\dfrac{2}{3}x+12[/tex]
Which implies that
[tex]x+y\geq x-\dfrac{2}{3}x+12=12+\dfrac{1}{3}x[/tex]
The mimum for this function, given that x ranges from 6 to 18, is clearly
[tex]12+\dfrac{1}{3}\cdot 6 = 12+2=14[/tex]
Case 1: x > 18
In this last case, any (positive) value for y is fine. Since x is at least 18, the sum of x and y will also be at least 18 (remember that y is non-negative).
So, out of all three cases, the minimum is achieved at x=6, y=8
