1. **Problem Statement:** We want to find the number of laptops $x$ to sell in order to maximize profit. Given: - Marginal cost function $MC(x)$ (not explicitly given, so we focus on revenue and cost). - Fixed cost = 500 (thousands of shillings). - Revenue $R(x)$ is quadratic: $R(x) = ax^2 + bx + c$. - Data points for revenue: - $R(10) = 1650$ - $R(25) = 3375$ - $R(40) = 4200$ 2. **Find the quadratic revenue function $R(x)$:** We have three equations: $$ \begin{cases} 100a + 10b + c = 1650 \\ 625a + 25b + c = 3375 \\ 1600a + 40b + c = 4200 \end{cases} $$ 3. **Solve the system:** Subtract first from second: $$ (625a - 100a) + (25b - 10b) + (c - c) = 3375 - 1650 \\ 525a + 15b = 1725 $$ Divide by 15: $$ 35a + b = 115 \quad (1) $$ Subtract first from third: $$ (1600a - 100a) + (40b - 10b) + (c - c) = 4200 - 1650 \\ 1500a + 30b = 2550 $$ Divide by 30: $$ 50a + b = 85 \quad (2) $$ Subtract (1) from (2): $$ (50a - 35a) + (b - b) = 85 - 115 \\ 15a = -30 \\ a = -2 $$ From (1): $$ 35(-2) + b = 115 \\ -70 + b = 115 \\ b = 185 $$ From first equation: $$ 100(-2) + 10(185) + c = 1650 \\ -200 + 1850 + c = 1650 \\ c = 1650 - 1650 = 0 $$ So, $$ R(x) = -2x^2 + 185x $$ 4. **Cost function:** Fixed cost = 500, marginal cost not given explicitly, so assume total cost $C(x) = 500 + MC(x)$. 5. **Profit function:** $$ P(x) = R(x) - C(x) = (-2x^2 + 185x) - (500 + MC(x)) $$ Since marginal cost is not given explicitly, assume it is constant or negligible for maximizing profit based on revenue. 6. **Maximize profit:** Maximize $P(x) = -2x^2 + 185x - 500$. Take derivative: $$ P'(x) = -4x + 185 $$ Set to zero: $$ -4x + 185 = 0 \\ x = \frac{185}{4} = 46.25 $$ 7. **Check closest options:** Options: 53, 27, 38, 42 Closest to 46.25 is 42. **Final answer:** $$\boxed{42}$$