A furniture trader deals in tables and chairs. He has ₹75,000 to invest and a space to store at most 60 items. A table costs him 1,500 and a chair costs him ₹1,000. The trader earns a profit of ₹400 and ₹250 on a table and chair, respectively. Assuming that he can sell all the items that he can buy, then which of the following is/are true for the above problem.
(A) Let the trader buys x tables and y chairs. Let Z denote the total profit. Thus, the mathematical formulation of the given problem is
Maximize $Z=400x +250y$
subject to constraints: $x + y ≤ 60, 3x + 2y ≤ 150, x ≥0, y ≥0$
(B) The corner points of the feasible region are (0, 0), (50, 0), (30, 30) and (0, 60).
(C) Maximum profit is ₹19,500 when trader purchase 60 chairs only
(D) Maximum profit is ₹20,000 when trader purchase 50 tables only
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B), and (D) only
Constraints:- $⇒x+y≤60$
$⇒1500x+1000y≤75000$
$⇒x≥0,y≥0$
Total profit function: Maximize $Z = 400x + 250y$
∴ (A) is true
(1) Intersection of $x+y=60$ & $3x+2y=150$
$y=60-x$
$3x+2(60-x)=150$
$x+120=150$
$x=30,y=60$
∴ One corner point is (30, 30)
(2) Intersection of $x+y=60$ with axis.
At $x=0,y=60→(0,60)$
At $y=0,x=60→(60,0)$ (not valid due to budget constraint)
(3) Intersection of $3x+2y$ with axis.
At $x=0,3(0)+2y=150,y=75$
$(0,75)$ is not valid due to constraints
At $y=0,3x+2(0)=150⇒x=50$
$(50,0)$ is valid
∴ Corner points of the feasible region are:
$(0,0),(50,0),(30,30),(0,60)$
∴ Statement B is correct.
At (0, 0): $Z=400(0)+250(0)=0$
At (50, 0): $Z=400(50)+250(0)=20,000$
At (30, 30): $Z=400(30)+250(30)=19,500$
At (0, 60): $Z=400(0)+250(60)=15,000$
∴ Maximum Profit is 20,000 when purchasing 50 tables.
∴ D is true and C is false.