A shopkeeper deals in two items-wall hangings and artificial plants. He has ₹15000 to invest and a space to store atmost 80 pieces. A wall hanging costs him ₹300 and an artificial plant ₹150. He can sell a wall hanging at a profit of ₹50 and an artificial plant at a profit of ₹18. Assuming that he can sell all the items that he buys, formulate a linear programming problem in order to maximize his profit.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Maximize $Z=50x+18y$, Subject to: $2x+y≤100$, $x+y≤80$, $x≥0, y≥0$
Let $x$ be the number of wall hangings and $y$ be the number of artificial plants that the dealer buys and sells. Then the profit of the dealer is $Z = 50x + 18y$, which is the objective function.
As a wall hanging costs ₹300 and an artificial plant costs ₹150, the cost of $x$ wall hangings and $y$ artificial plants is $300x + 150y$. We are given that the dealer can invest atmost ₹15000. Hence, the
investment constraint is
$300x + 150y ≤ 15000$ i.e. $2x + y ≤ 100$
As the dealer has space to store atmost 80 pieces, we have another constraint (space constraint):
$x + y ≤80$
Also the number of wall hangings and artificial plants can't be negative. Thus we have the non-negativity constraints:
$x≥0, y ≥0$.
Thus the mathematical formulation of the L.P.P. is:
Maximize $Z = 50x + 18y$ subject to the constraints
$2x + y ≤ 100, x + y ≤80, x≥0, y ≥0$.