question_answer

The maximum value of \[z=2x+5y\] subject to the constraints \[2x+5y\le 10,x+2y\ge 1,x-y\le 4,x\ge y\ge 0,\] Occurs at

A) Exactly one pint

B) Exactly two points

C) Infinitely many points

D) None of these

Correct Answer: C

Solution :

[c] We find that the feasible region is on the same side of the line \[2x+5y=10\]as the origin, on the same side of the line \[x-y=4\]as the origin and on the opposite side of the line \[x+2y=1\]from the origin. Moreover, the lines meet the coordinate axes at (5, 0), (0, 2); (1, 0), (0, 1/2) and (4, 0). The lines \[x-=4\]and \[2x+5y=10\]intersect at\[\left( \frac{30}{7},\frac{2}{7} \right)\]

The values of the objective function at the vertices of the pentagon are:

(i) \[Z=0+\frac{5}{2}=\frac{5}{2a}\]

(ii) \[Z=2+0=2\]

(iii) \[Z=8+0=8\]

(iv) \[Z=\frac{60}{7}+\frac{10}{7}=10\]

(v) \[Z=0+10=10\]

The maximum value 10 occurs at the points D (30/7, 2/7) and E (0, 2). Since D and E are adjacent vertices, the objective function has the same maximum value 10 at all the points on the lines DE.