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NCERT Exemplar · Q42

Q.Refer to Exercise 30. Minimum value of FF is
(A) 00
(B) −16-16
(C) 1212
(D) does not exist

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Evaluating F=3x−4yF = 3x - 4y at the corner points of the Exercise-30 feasible region gives 00, 1212 and −16-16. Since FF is bounded on this region, the minimum exists and equals the least corner value, −16-16 — option (B).

Set-up

Exercise 30 gives the objective F=3x−4yF = 3x - 4y over the shaded feasible region of Fig. 12.10, whose corner points are (0,0)(0,0), (12,6)(12,6) and (0,4)(0,4) (with x≥0x \ge 0, y≥0y \ge 0). We use the graphical corner-point method: evaluate FF at every corner.

Corner-point method

Corner pointF=3x−4yF = 3x - 4y
(0,0)(0,0)3(0)−4(0)=03(0) - 4(0) = 0
(12,6)(12,6)3(12)−4(6)=123(12) - 4(6) = 12
(0,4)(0,4)3(0)−4(4)=−163(0) - 4(4) = -16

These three values — 00, 1212 and −16-16 — are exactly the numeric options offered.

Does the minimum exist? …

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