A company makes two types of cakes: type A and type B. It costs Rs. 360 to make a type A cake and Rs. 120 to make a type B cake. The company can make at most 300 cakes and spend at most Rs. 72,000 a day, the number of cakes of type B \( (y) \) cannot exceed the number of type A \( (x) \) cakes by more than 200. The company makes a profit of Rs. 200 on each type A cake and Rs. 50 on each type B cake. A LPP to maximize the profit (Rs.) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Maximize \( Z=200x+50y \) subjected to \( 3x+y\le 600,x+y\le 300 \) \( y\le x+200,x\ge 0,y\ge 0 \)
Let x and y be the numbers of type A and type B cakes made in a day.
Profit: \( Z=200x+50y \), to be maximised.
Cost: at most Rs. 72,000 a day:
\( \displaystyle 360x+120y\le 72000\ \Rightarrow\ 3x+y\le 600 \)
Number of cakes: at most 300:
\( \displaystyle x+y\le 300 \)
y cannot exceed x by more than 200, that is, \( y-x\le 200 \):
\( \displaystyle y\le x+200 \)
And the numbers of cakes cannot be negative: \( x\ge 0,\ y\ge 0 \).
These are exactly the constraints in Option (2).
Why the other options are incorrect
Option (1) Maximize \( Z=200x+50y \) subjected to \( 3x+y\le 600,x+y\le 300 \) \( y\ge x+200,x\ge 0,y\ge 0 \) — Incorrect. \( y\ge x+200 \) says B cakes exceed A cakes by at least 200, the opposite of “cannot exceed by more than 200”.
Option (3) Maximize \( Z=200x+50y \) subjected to \( 3x+y\ge 600,x+y\ge 300 \) \( y\ge x+200,x\ge 0,y\ge 0 \) — Incorrect. All three inequalities are reversed: “at most” means ≤, and y − x must be at most 200.
Option (4) Maximize \( Z=200x+50y \) subjected to \( 3x+y\le 600,x+y\ge 300 \) \( y\le x+200,x\ge 0,y\ge 0 \) — Incorrect. \( x+y\ge 300 \) means at least 300 cakes, but the company can make at most 300.