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Use graphical methods to solve the linear programming problem.

Maximize: $z= 4x+2y$

subject to : $x-y\le 7$

$19x+12y\le 228$

$18x+18y \le 324$

$x\ge 0,y\ge 0$

the max is ?? when x= ?? and y=???

my effort

$x-y\le 7$

when $x= 0 , y=0$,we get,$(0, -7) ,(7, 0)$

$19x+12y\le 228$

when $x= 0 , y=0$,we get $(12, 0) ,(0, 19)$

$18x+18y \le 324$

when$ x= 0 , y=0$, we get, $ (18, 0),(0, 18)$

We evaluate the objective function to maximize at each corner point:

$z= 4x+2y$

$(0,0)\Rightarrow 0$

$(7, 0)\Rightarrow 28$

$(12, 0)\Rightarrow 48$

$(18, 0)\Rightarrow 72$

The maximum value of $z$ is $72$, and will occur at $(18,0)$, when $x=18$ and $y=0$

Please review my answer.

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  • $\begingroup$ meta.math.stackexchange.com/questions/5020/… $\endgroup$ Commented Jul 30, 2015 at 17:08
  • $\begingroup$ You forgot the constraint $x-y \leq 7$, which is not satisfied by your solution. Perhaps you took the wrong side of the constraint? $\endgroup$ Commented Jul 30, 2015 at 17:12
  • $\begingroup$ how that ??????? $\endgroup$ Commented Jul 30, 2015 at 17:12
  • $\begingroup$ Pretty straightforward, isn't it? $x-y = 18-0 = 18 > 7$. $\endgroup$ Commented Jul 30, 2015 at 17:13
  • $\begingroup$ You should draw out the constraint set and examine which points are actually in it. $\endgroup$ Commented Jul 30, 2015 at 17:14

1 Answer 1

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Try plotting the region using WolframAlpha

(The code looks like this: RegionPlot[x <= 7 + y && 19 x + 12 y <= 228 && x + y <= 18, {x, -10, 10}, {y, -20, 30}])

It has 4 vertices, figure out where is the extreme value of the objective function you are looking for...

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