Minimize Z = 2x+3y,subject to the constraints x?0,y?0,x+2y ?1 and x+2y?10 The answer given in the book is :Minimum Z = 3/2 at(0,1/2)
Asked by Manoj
| 20th Jun, 2013,
12:08: PM
Expert Answer:
if you draw a graph for the given 4 constraint equation -- x?0,y?0,x+2y ?1 and x+2y?10 and identify the polygon enclosed by the equations, you will find that its is a quadrilateral, with A(1,0); B(10,0); C(0,5); D(0,1/2)
Now since its an enclosed area, the minimum value of Z should be at one of the vertices.
hence, at A(1,0); Z = 2(1) + 3(0) = 2
at B(10,0); Z = 2(10) + 3(0) = 20
at C(0,5); Z = 2(0) + 3(5) = 15
at D(0,1/2); Z = 2(0) + 3(1/2) = 3/2
So, the minimum value of Z is at C(0,1/2) and the value is 3/2.
if you draw a graph for the given 4 constraint equation -- x?0,y?0,x+2y ?1 and x+2y?10 and identify the polygon enclosed by the equations, you will find that its is a quadrilateral, with A(1,0); B(10,0); C(0,5); D(0,1/2)
Now since its an enclosed area, the minimum value of Z should be at one of the vertices.
hence, at A(1,0); Z = 2(1) + 3(0) = 2
at B(10,0); Z = 2(10) + 3(0) = 20
at C(0,5); Z = 2(0) + 3(5) = 15
at D(0,1/2); Z = 2(0) + 3(1/2) = 3/2
So, the minimum value of Z is at C(0,1/2) and the value is 3/2.
Answered by
| 23rd Jun, 2013,
07:39: AM
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