Determine the maximum value of $Z=3x+4y$ if the feasible region (shaded) for a LPP is shown in given figure.
Answer & explanation
Correct answer: option 3
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.
|
Corner points |
Corresponding value of $Z=3x+4y$ |
|
(0, 0) |
0 |
|
(52, 0) |
156 |
|
(44, 16) |
196 → Maximum |
|
(0, 38) |
152 |
Hence, Z is at (44, 16) is maximum and its maximum value is 196.