`2x+3yle35,yge2,xge5`

2x+3yle6,x+4yle4,xge0,yge0

x-2yge0,2x-yle-2,xge0,yge0

The corner points of the feasible region determined by the following sytem of linear inequalities: 2x+yle10, x+3yle15, x, yge0 are (0,0),(5,0),(3,4) and (0,5). Let Z=px+qy where p,qge0 . Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is

If the function f: R to R is defined by f (x) {:{( 2x+1 ,xlt-1) ,( x^(2) -1,le x lt 3 ),( 3x-2, xge 3) :} "Find " (f(-1) +f(3))/( 2)

Consider the linear inequalities 2x+3yle6, 2x+y le 4, x ge0, y ge0 (i) Mark the feasible region. (ii) Maximise the function Z=4x+5y subject to the given constraints.

Minimise Z=3x+2y Subject to the constraints x+yge8 3x+5y le 15 x ge0, y ge0

Determine graphically the minimum value of the objective function Z=-50x+20y ……………1 subject to the constraints: 2x-yge-5 …………..2 3x+yge3 ………….3 2x-3yle12 ……………..4 xge0, yge0 …………5

f(x)={{:(x^(2)", "xle2),(8-2x", "2ltxlt4),(4", "xge4):}