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16. [6 points] Solve the linear programming problem below graphically:PageMaximize M = 4x1 + 3x2 subject to X1 + Zxz < 10 2x1 + Xz < 12 X1+ Xz < 7Xv' ...

Question

16. [6 points] Solve the linear programming problem below graphically:PageMaximize M = 4x1 + 3x2 subject to X1 + Zxz < 10 2x1 + Xz < 12 X1+ Xz < 7Xv' Xz 2 0

16. [6 points] Solve the linear programming problem below graphically: Page Maximize M = 4x1 + 3x2 subject to X1 + Zxz < 10 2x1 + Xz < 12 X1+ Xz < 7 Xv' Xz 2 0



Answers

Solve each linear programming problem. $$ \text { Maximize } z=2 x+y \quad \text { subject to } \quad x \geq 0, \quad y \geq 0, \quad x+y \leq 6, \quad x+y \geq 1 $$

Okay. So in this linear programming problem, we're given our objective function as well as our constraints. And we want to graft those constraints. And we know that because X and Y are both greater than or equal to zero, it occurs in quadrant one, and I want to graft these two functions. But first I'm going to write them out so that why is less than or equal to negative X plus six. I'm just rearranging the equations. Do you have X Plus One? This makes it easier for me to graph it. And we know that our Y intercept for our first equation is at six. There's a slope of negative one ending at the X intercept of six and the same thing for the second equation, except at the Y and exit intercepts of one and are feasible. Points occur in between our constraints. And now we want to list the corner points to go on to the next step. So our corner points, as you can see here, are zero comma. One appear is zero comma. Six down here would be one common zero, and down here would be six comma zero And now what we want to do is plug those into our objective function and self busy for each corresponding point. And so when I plug zero comma one inter objective function, I end up with Z equals one 06 I end up with is equal six 10 I get Z equals two. And for six year I end up with Z equals 12. And the problem asked us to find the maximum value. So look for the highest number and we know that that's 12. And so we end up with the maximum value of Z is 12 and it occurs at the 120.6 comma zero.

In order to maximize the objective function, D equals three X plus five y. First, we have to grab the constraints to which it is subject to X and Y. Being greater than or equal to zero puts us in the first quadrant and the other constraints are graft and corresponding colors. From these constraints, we can find feasible points at the following circle locations. These are the points corresponding to the circle locations. In order to find out which one of these points maximizes he we have to plug in each point into the objective function. Z equals three X plus five y Doing so gives us these values. From these we can see that the point that maximizes e is zero for which gives us a value for Z of 20.

In order to maximize the objective function, C equals X plus three y. We have to graft the constraints. It is subject to X and Y being greater than or equal to zero puts us in the first quadrant and the other constraints are graft and corresponding colors. These constraints give us feasible points at the following circled locations. These are the points that have been circled. In order to find out at which one of these points Z is maximized, you have to plug in each point for X and Y into the objective function. Doing so gives you the following values. From these, we can see that at the 0.57 z is maximized at a value of 26.

In order to maximize the objective function, C equals five x plus three y. We have to graft the constraints to which it is subject to X and Y being greater than or equal to zero puts us in the first quadrant and the other constraints are graft and their corresponding colors. From these constraints, we confined the feasible points at the following circle locations. These are the points corresponding to the circled locations. In order to find which one of these points maximizes Z equals five x plus three y, we will plug in each one of these points in for X and y Doing so gives us the following values from these values we confined that the point that maximizes he is to six, giving us a value of 28 for Z.


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