The points you found are the vertices of the feasible region. I agree with the first three points you got. However, the last point should be (25/11, 35/11). This point is at the of the intersection of the two lines 8x-y = 15 and 3x+y = 10
So the four vertex points are: (1,9) (1,7) (3,9) (25/11, 35/11)
Plug each of those points, one at a time, into the objective function z = 7x+2y. The goal is to find the largest value of z
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Plug in (x,y) = (1,9) z = 7x+2y z = 7(1)+2(9) z = 7+18 z = 25 We'll use this value later. So let's call it A. Let A = 25
Plug in (x,y) = (1,7) z = 7x+2y z = 7(1)+2(7) z = 7+14 z = 21 Call this value B = 21 so we can refer to it later
Plug in (x,y) = (3,9) z = 7x+2y z = 7(3)+2(9) z = 21+18 z = 39 Let C = 39 so we can use it later
Finally, plug in (x,y) = (25/11, 35/11) z = 7x+2y z = 7(25/11)+2(35/11) z = 175/11 + 70/11 z = 245/11 z = 22.2727 which is approximate Let D = 22.2727
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In summary, we found A = 25 B = 21 C = 39 D = 22.2727
The value C = 39 is the largest of the four results. This value corresponded to (x,y) = (3,9)
Therefore the max value of z is z = 39 and it happens when (x,y) = (3,9)
2/3 also = 4/6. 5/8 also = 10/16 hope this helps xx just multiply fractions by same number. as long as answer is reducible to original fraction, should be correct
