Consider a L.P.P, Maximize $Z=3x+5y$ subject to constraints $2x+6y ≤ 6,x-y≥0,x≥0, y≥0$. Then which of the following are true?
(A) The feasible region of L.P.P is bounded region
(B) The corner points of the feasible region are (0, 0), (2, 2), (0, 1)
(C) Maximum value of Z is 9
(D) Point (1, 3) lies in the feasible region
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (C) only
Maximize $Z=3x+5y$ subject to
$2x+6y\le 6,\; x-y\ge 0,\; x\ge 0,\; y\ge 0$
Simplify the first constraint: $x+3y\le 3$.
Find corner points of the feasible region (intersection of boundary lines):
Intersection of $x+3y=3$ and $x=y$:
$x+3x=3\Rightarrow 4x=3\Rightarrow x=\frac{3}{4},\; y=\frac{3}{4}$
Intersection of $x+3y=3$ with $y=0$:
$x=3\Rightarrow (3,0)$
Intersection of $x=y$ with $y=0$ gives $(0,0)$.
Thus corner points: $(0,0),\;\left(\frac{3}{4},\frac{3}{4}\right),\;(3,0)$.
Evaluate $Z$ at these points:
At $(0,0):\;Z=0$
At $\left(\frac{3}{4},\frac{3}{4}\right):\;Z=3\left(\frac{3}{4}\right)+5\left(\frac{3}{4}\right)=\frac{24}{4}=6$
At $(3,0):\;Z=3\cdot 3+5\cdot 0=9$
Maximum value of $Z$ is $9$ at $(3,0)$. The feasible region is bounded. Point $(1,3)$ is not feasible since $1+3\cdot 3=10>3$.
Correct statements: (A) True, (B) False, (C) True, (D) False.