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

1. Draw all graphs and determine the shaded region:

a) for you should take 1 and 4 quadrants (including y-axis);

b) for you should take 1 and 2 quadrants (including x-axis);

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

d) for 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.

Maximum value is 15 at point (3,0)

you should at least use a calculater

step-by-step explanation:

c

step-by-step explanation:

at least i think its