CP = 80 × 40
Profit from the n objects = n% × 40 × n.
Profit from the remaining objects = (100 – n)% × 40 × (80 – n).
We need to find the minimum possible value of n% × 40 × n + (100 – n)% × 40 × (80 – n).
Or, we need to find the minimum possible value of n^{2} + (100 – n) (80 – n).
Minimum of n^{2} + n^{2} – 180n + 8000
Minimum of n^{2} – 90n + 4000
Minimum of n^{2} – 90n + 2025 – 2025 + 4000
We add and subtract 2025 to this expression in order to crate an expression that can be expressed as a perfect square. This approach is termed as the “Completion of Squares” approach. We keep revisiting this in multiple chapters.
Minimum of n^{2} – 90n + 2025 + 1975 = (n – 45)^{2} + 1975
This reaches minimum when n = 45.
When n = 45, the minimum profit made
45% × 40 × 45 + 55% × 40 × 35
18 × 45 + 22 × 35 = 810 + 770
Rs. 1580. Answer choice (C)
Correct Answer: Rs. 1580