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NCERT Exemplar · Q44

Q.Refer to Exercise 32. Maximum of FF - Minimum of F=F =
(A) 6060
(B) 4848
(C) 4242
(D) 1818

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Evaluating F=4x+6yF=4x+6y at the corner points gives a maximum of 7272 and a minimum of 1212, so Max −- Min =60=60 — option (A).

What the question refers to

This item follows on from the earlier exercise whose feasible region has the corner points

(0,2),(3,0),(6,0),(6,8),(0,5),(0,2),\quad (3,0),\quad (6,0),\quad (6,8),\quad (0,5),

with objective function F=4x+6yF = 4x + 6y. By the Corner-Point Theorem, the largest and smallest values of a linear objective over a feasible region are always found at these vertices, so we simply test each one.

Evaluate F at every corner

Corner (x,y)(x,y)F=4x+6yF = 4x + 6y
(0,2)(0,2)0+12=120+12 = 12
(3,0)(3,0)12+0=1212+0 = 12
(6,0)(6,0)24+0=2424+0 = 24

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