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3 years ago

Determine the maximum value of the objective function, P.

Mathematics

1 answer:

Ad libitum [116K]3 years ago

5 0

Answer:

Maximum (250,125) Answer

Step-by-step explanation:

This is more of a graphing problem than it is anything else.

Begin by graphing all 4 given equations.

When you do that, mark the intersection points of at least 2 lines. In this case it is exactly 2 lines for each intersecting point.

6x + 4y <= 2000 and 2x + 4y <= 1000 intersect at (250,125)

x=>0 and y=>0 intersect at (0,0)

6x + 4y <=2000 and x => 0 intersect at 333.333

2x + 4y <=1000 and y>=0 intersect at 0,250.

Any other intersection points fall outside the range of the givens. The shaded part we are interested in is sort of a very dark green/blue. It is the interior of the quadrilateral determined by the 4 vertices that are marked.

Now all you have to do is determine the maximum point using P=15x + 12y

For 0,0 P = 15*0 + 12,0 = 0

For 0,250 P = 15*0 + 12*250 = 3000

For 333.3333,0 P = 15*333.3333 + 12*0 = 5000 rounded.

For 250,125 P = 15*250 + 12*125 = 5250 Which is the maximum

Answer (250,125) produces the maximum value Answer

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<h3><u>Explanation</u></h3>

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Answer Check by substituting both x and y values in both equations.

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$6x - 2y = 10 \\ 6(3) - 2(4) = 10 \\ 18 - 8 = 10 \\ 10 = 10 \longrightarrow \sf{true} \: \green{ \checkmark}$

<u>Second</u><u> </u><u>Equation</u>

$x - 2y = - 5 \\ 3 - 2(4) = - 5 \\ 3 - 8 = - 5 \\ - 5 = - 5 \longrightarrow \sf{true} \: \green{ \checkmark}$

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<h3><u>Answer</u></h3>

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