A bakery bakes x cakes and y loaves of bread a day. Its oven can hold at most 20 items a day. A cake uses 3 units of flour and a loaf uses 1 unit, and 36 units of flour are available a day. The profit is Rs. 50 on a cake and Rs. 30 on a loaf. The daily plan that gives the maximum profit is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8 cakes and 12 loaves
The problem is
Maximise \(Z = 50x + 30y\)
subject to
\(x + y \le 20\) (oven)
\(3x + y \le 36\) (flour)
\(x \ge 0,\ y \ge 0\)
Subtracting the first line from the second,
\(2x = 16\)
\(x = 8,\ y = 12\)
The corner points are (0, 0), (12, 0), (8, 12) and (0, 20). Values of Z:
\(Z(0, 0) = 0\)
\(Z(12, 0) = 600\)
\(Z(8, 12) = 400 + 360 = 760\)
\(Z(0, 20) = 600\)
The optimal solution is (8, 12): 8 cakes and 12 loaves, with a profit of Rs. 760.
Why the other options are incorrect
Option (1) 12 cakes and no loaves — Incorrect. It is feasible, but the profit is only Rs. 600.
Option (2) no cakes and 20 loaves — Incorrect. It is feasible, but the profit is only Rs. 600.
Option (3) 10 cakes and 10 loaves — Incorrect. It needs \(30 + 10 = 40 \gt 36\) units of flour, so it is not feasible.