The corner points of the bounded feasible region determined by \( x+3y\le 60,x+y\ge 10,x\le y,x\ge 0,y\ge 0 \) are A(0, 10), B(5, 5), C(15, 15) and D(0, 20). If the objective function \( z=ax+by \) has its maximum value on the line segment CD, then the relation between \( a \) and \( b \) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 3a=b \)
The maximum occurs at every point of the segment CD only when z has the same value at both end points C and D (and that common value is the largest among the corners).
\( \displaystyle z(C)=15a+15b \)
\( \displaystyle z(D)=0\cdot a+20b=20b \)
\( \displaystyle 15a+15b=20b \)
\( \displaystyle 15a=5b\ \Rightarrow\ b=3a \)
This common value is the maximum: with b = 3a and a > 0 (positive coefficients, as usual for such a question),
\( \displaystyle z(A)=10b=30a \)
\( \displaystyle z(B)=5a+5b=20a \)
\( \displaystyle z(C)=z(D)=60a \)
60a is the largest, so the maximum is attained all along CD when 3a = b.
NCERT Exemplar:
Linear Programming
