The maximum value of the objective function $Z = 10x + 15y$ of an L.P.P, subjected to the constraints $2x + 4y ≤ 8, 3x + y ≤6, -x-y≥-4, x≥0, y ≥ 0$ is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 34
$Z=10x+15y.$
$2x+4y\le8 \Rightarrow x+2y\le4.$
$3x+y\le6.$
$-x-y\ge-4 \Rightarrow x+y\le4.$
$x\ge0,\;y\ge0.$
$\text{Corner points of feasible region:}$
$(0,0).$
$(2,0)\;(\text{from }3x+y=6,\;y=0).$
$(0,2)\;(\text{from }x+2y=4).$
$(1.6,1.2)\;(\text{intersection of }x+2y=4 \text{ and }3x+y=6).$
$Z(0,0)=0.$
$Z(2,0)=20.$
$Z(0,2)=30.$
$Z(1.6,1.2)=10(1.6)+15(1.2)=34.$
$\text{Maximum value of }Z=34.$