Question

Maximize z = 2x + 4y subject to the constraints

x ≥ 0, y ≥ 0, x + y ≥ 1, 3x + 2y <= 6

Answer 1

$x\le 0$

When $x$ is greater than zero it mean tangent of that mean the feasible region is on or to the right of the line where equation is zero$(x=0)$ which is the line which is $y$ axis

$y\le 0$

Where y is greater that it means above that mean feasible region is on or above the line where equation is $y=0$ which is the line which is x-axis

Those two mean that the feasible region is in the upper right hand part of the xy co ordinate system.

See image 1

When x is greater that it mean right of

$y\le 1-x$

That mean feasible region os on or of above the line where equation is $x+y=1$

See image 2

$3x+2y\le 6$

If we solve

$x\le \frac{6-2y}{3}y\le \frac{6-3x}{2}$

$xintercept(2,0)yintercept(0,3)Cornerpointvalueof2=2x+4y$

$(1,0)2=2\left(1\right)+4\left(10\right)=2+0=2(2,0)2=2\left(2\right)+4\left(0\right)=4+0=4(0,3)2=2\left(0\right)+4\left(3\right)=0+12=12(0,1)2=2\left(0\right)+4\left(1\right)=0+4=4$

Maximum value when $x=0y=3$