Determine the maximum value of $Z=11x+7y$ subject to the constraints: $2x + y ≤6,x≤2,x≥0,y≥0$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 42
Given that: $Z=11x+7y$ and the constraints $2x+y≤6,x≤2, x≥0,y≥0$.
Let $2x+y=6$
|
x |
0 |
3 |
|
y |
6 |
0 |
The shaded area OABC is the feasible region determined by the constraints $2x+y≤6,x≤2,x≥0, y≥0$.
The feasible region is bounded.
So, maximum value will occur at a corner point of the feasible region.
Corner points are (0, 0), (2, 0), (2, 2) and (0,6).
Now, evaluating the value of Z, we get
|
Corner points |
Value of Z |
|
O(0, 0) |
$11(0)+7(0)=0$ |
|
A(2, 0) |
$11(2)+7(0)=22$ |
|
B(2, 2) |
$11(2)+7(2)=36$ |
|
C(0, 6) |
$11(0)+7(6)=42$ ← Maximum |
Hence, the maximum value of Z is 42 at (0, 6).