Question

What point in the feasible region maximizes the objective function constraints: x>=0 y>=0 -x+3>=y y<=1/3 x+1 Objective function: C=5x-4y

Answer 1

We are given with
Objective function
C = 5x - 4y
x ≥ 0
y ≥ 0
-x + 3 ≥ y
y ≤ (1/3)x + 1
The following points enclose the feasible region
(0. 0)
(0, 3)
(0,1)
(2/3,2/3)

Answer 2

1. Draw all graphs and determine the shaded region:

a) for $x\ge 0$ you should take 1 and 4 quadrants (including y-axis);

b) for $y\ge 0$ you should take 1 and 2 quadrants (including x-axis);

c) for $-x+3\ge y$ you should draw the graph of the line y=-x+3 and take the bottom half-plane;

d) for $y\le \dfrac{1}{3} x+1$ you should draw the graph of the line y=1/3x+1 and take the bottom half-plane.

Then the intersection of all shaded parts is needed shaded region, quadrilateral OABC (see diagram for details).

2. Th graphs of the lines C=5x-4y will intersect this shaded region at different points. The maximum of the function could be only at vertices.

3. Find the coordinates of all vertices:

a) O(0,0) - the origin;

b) A(0,1) - the y-intercept of line y=-x+3;

c) C(3,0) - the x-intercept of line y=1/3x+1;

d) B: solve the system y=1/3x+1, y=-x+3.

-x+3=1/3x+1,

-x-1/3x=1-3,

-4/3x=-2,

x=3/2,

y=-3/2+3=3/2,

then B(3/2,3/2).

4) F(x)=5x-4y and

F(O)=0,

F(A)=5·0-4·1=-4,

F(C)=5·3-4·0=15,

F(B)=5·3/2-4·3/2=7.5-6=1.5

The maximal value of F is at point C.

Answer: Maximum value is 15 at point (3,0)