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Miscellaneous Exercise 7 · Q57

Q.The corner points of the feasible solution are (0, 0), (2, 0), (127,37)\left(\dfrac{12}{7}, \dfrac{3}{7}\right), (0, 1). Then Z=7x+yZ = 7x + y is maximum at _______. A) (0, 0)
B) (2, 0)
C) (127,37)\left(\dfrac{12}{7}, \dfrac{3}{7}\right)
D) (0, 1)

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Evaluating Z=7x+yZ=7x+y at each given corner point: (0,0)→0(0,0)\to0; (2,0)→14(2,0)\to14; (127,37)→7(127)+37=12+37=877≈12.43\left(\tfrac{12}{7},\tfrac37\right)\to7\left(\tfrac{12}{7}\right)+\tfrac37=12+\tfrac37=\tfrac{87}{7}\approx12.43; $(0, …

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