Determine the maximum value of $Z=3x+4y$ if the feasible region (shaded) for a LPP is shown in given figure.

196

The correct answer is Option (3) → 196

Identify the Corner Points

The corner points are the vertices of the shaded region:

• O: Origin $(0, 0)$
• A: $x$-intercept of the line $2x + y = 104$
• Set $y = 0 ⇒2x = 104 ⇒x = 52$. Point A is $(52, 0)$.
• D: $y$-intercept of the line $x + 2y = 76$
• Set $x = 0 ⇒2y = 76 ⇒y = 38$. Point D is $(0, 38)$.
• E: The intersection of lines $2x + y = 104$ and $x + 2y = 76$.
• From the second line, $x = 76 - 2y$.
• Substitute into the first: $2(76 - 2y) + y = 104 ⇒152 - 4y + y = 104 ⇒-3y = -48 ⇒y = 16$.
• Substitute back: $x = 76 - 2(16) = 76 - 32 = 44$. Point E is $(44, 16)$.

Here, $2x+y=104$ and $2x+4y=152$ intersect at $⇒E(44,16)$.

As clear from the graph, corner points are $O,A,E$ and $D$ with coordinates (0, 0), (52, 0), (144, 16) and (0, 38), respectively. Also, given region is bounded.

Hence, Z is at (44, 16) is maximum and its maximum value is 196.