Paint Mixture Profit Maximization

Total Selling Price = Rs. 264 Profit = 10% Total Cost Price = $264 / 1.1 = 240$ Average cost per litre = $240 / 10 = 24$

Let cost of paint B be $x$. Then cost of paint A is $x + 8$. Since the quantity of B does not exceed A, the proportion of paint A is at least 50% ($A \ge B$). Average cost = $x + 8 \cdot \frac{A}{A + B} = 24$ $\Rightarrow 8 \cdot \frac{A}{10} = 24 - x$ Since $A \ge 5$, $A/10 \ge 0.5$. Maximum value of $x$ occurs when $A$ is minimized, i.e., $A = 5$. $8 \cdot 0.5 = 24 - x \Rightarrow 4 = 24 - x \Rightarrow x = 20$.