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2ueiFor the zbove L P prablem answer the following questions here; You must show all of vour wcrk on the scrap pages to jecelve full crcditWhat Is the ontmal soluti...

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2ueiFor the zbove L P prablem answer the following questions here; You must show all of vour wcrk on the scrap pages to jecelve full crcditWhat Is the ontmal solutinn? Wnite J5 an nrdered nalr; Use decimal only necdedIb) What Is the vajue of the objective function at the optimal solution? Use decimal places(c) Which constraints are binding? Use constralnt numnbersUI wholls lhe Slack/Surplus for thc nen-binding constrslntfs) Intcrprct I le; SO hours ot fnishing timc, lock.

2uei For the zbove L P prablem answer the following questions here; You must show all of vour wcrk on the scrap pages to jecelve full crcdit What Is the ontmal solutinn? Wnite J5 an nrdered nalr; Use decimal only necded Ib) What Is the vajue of the objective function at the optimal solution? Use decimal places (c) Which constraints are binding? Use constralnt numnbers UI wholls lhe Slack/Surplus for thc nen-binding constrslntfs) Intcrprct I le; SO hours ot fnishing timc, lock.



Answers

Find the indicated maximum and minimume values by the linear programming method of this section. For Exercises $5-16,$ the constraints are shown below the objective function. Maximum $P:$ $P=8 x+6 y$ $x \geq 0, y \geq 0$ $2 x+5 y \leq 10$ $4 x+2 y \leq 12$

Problem seven. We want to get the maximum B for this objective function, knowing that these are the constraints of the problem. First, let's draw the constraints to find the visible points for our problems. We have your Y. You have your ex. The first concern is our access because we want to get both the values of X. Then we have this area. The second constraint is the X axis. And we want the values above the X axis, positive values avoid. And the third constraint we should draw the line. X plus two. White equal six. When X equals zero, we have Y equals three. Have a point here and when? Why equal zero have X equals six. Then we have a point here. We draw the line and this the origin for this inequality. We can see that zero is smaller than equal sex. It satisfies the inequality. Then we have the area with the origin, the area that towards the origin. Which means we have the visible area, this area and the physical points at this point. This point this point let's substitute by these points in the objective function to get the maximum peak, the first point X equals zero, Y equals three, then B equals nine, multiplied by three gifts 27. Second point is zero and zero, then B equals zero. The third point is six and zero, Then it equals 30 five, multiplied by six gifts 30. And this is the maximum B, the maximum equals 30 which is the final answer of our problem.

In problem six. We want to get the maximum p. For the objective function B equals two. X plus seven. One. And these are the constraints. First let's draw the visible points. Yeah. X and Y. The first constraint is X is greater than equals one. Then we have the line X equals one this line and we have the points to the right because we want X to be greater than one and for Y is greater than zero, has the points above the X axis. For the third constraint we draw the line X plus four. Y equals it. When you substitute X equals zero, We get what equals two. Good point here. And when we substitute y equals zero. Get x equals it. 12345678 have a point here. Then this is the line. Let's just the origin. When we substitute by the origin we got zero plus zero, smaller than equals it. Then it's his voice. The inequality which means we have the area that points to the origin. Then the physical area is this triangle. And the physical points is this point this point at this point let's get the values of P. At these points we have X. Y. P. The first point Has x equals zero. Sorry, at x equals one. And why? Let's substitute in this line by X equals want to get why When x equals one We can get y equal seven divided by four. And it's seven divided by four and be equals two. X Plus seven multiplied by boy. Then it's 2-plus 49 divided by four. It's 14 0.25 For the second point we have X equals one. I equals zero. It's too third point we have X equals it. Y equals zero. It gives It multiplied by two f. 16. This is the maximum be, then the maximum equals 16. This is the final answer of our problem.

In the problem, this is the desired car for the and gasoline with respect to time. So we have this as quantity that is milligram and time in days, so this is the car, and it is the answer to the problem.

In problem for the constraints of millennial programming problem Has the physical points of one and 3. Eight and zero, nine and seven. Five and eight. Zero and six. And finally the same 0.1 and three. Again, we want to get the maximum and minimum values of the objective function F equals two X plus five boy in this region. To do so we substitute by each point. In the physical region. We substitute only by the physical points and F here and get the maximum and minimum values. Let's try the first point. We have to buy buy one plus five, multiplied by three. If 17 the second point is two, multiplied by eight plus zero gives 16. Third point is two, multiplied by nine Plus five months. Applied by seven gives 53. Third point gives two, multiplied by five Plus five, multiplied by it gives 50. And the last point gibbs five to multiply by six plus zero gives 30. We can find the minimum value is here. If minimum equals 16 and the maximum value is here then if maximum equals 53. And this is the final answer of our problem.


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