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If the objective function Z=px+qy,p>0,q>0Z = px + qy, p > 0, q > 0 of a linear programming problem attains its optimal value at the points (4, 7) and (5, 5) and pq=50pq = 50 then

Solution

✅ Correct Option: 2

The objective function Z=px+qyZ = px + qy where p>0,q>0p > 0, q > 0 attains its optimal value at both points (4,7)(4, 7) and (5,5)(5, 5).

Since the optimal value occurs at two different points, both points give the same value of ZZ.


At point (4,7)(4, 7): Z=4p+7qZ = 4p + 7q

At point (5,5)(5, 5): Z=5p+5qZ = 5p + 5q

Setting them equal:

4p+7q=5p+5q4p + 7q = 5p + 5q

7q−5q=5p−4p7q - 5q = 5p - 4p

2q=p2q = p


Given that pq=50pq = 50, substitute p=2qp = 2q:

(2q)×q=50(2q) \times q = 50

2q2=502q^2 = 50

q2=25q^2 = 25

q=5q = 5 (taking the positive value since q>0q > 0)


Using p=2qp = 2q:

p=2(5)p = 2(5)

p=10p = 10


Therefore, p=10p = 10 and q=5q = 5.

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