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, but assumed known or derivable). - Fixed cost = 500 (thousands of shillings). - Total 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. **Step 1: Find the quadratic revenue function $R(x)$** We use the general form $R(x) = ax^2 + bx + c$ and plug in the data points: $$ \begin{cases} 100a + 10b + c = 1650 \\ 625a + 25b + c = 3375 \\ 1600a + 40b + c = 4200 \end{cases} $$ 3. **Step 2: Solve the system of equations** Subtract the first from the second: $$625a - 100a + 25b - 10b + c - c = 3375 - 1650$$ $$525a + 15b = 1725$$ Divide by 15: $$35a + b = 115 \\ (1)$$ Subtract the first from the third: $$1600a - 100a + 40b - 10b + c - c = 4200 - 1650$$ $$1500a + 30b = 2550$$ Divide by 30: $$50a + b = 85 \\ (2)$$ Subtract (1) from (2): $$(50a + b) - (35a + b) = 85 - 115$$ $$15a = -30$$ $$a = -2$$ Plug $a = -2$ into (1): $$35(-2) + b = 115$$ $$-70 + b = 115$$ $$b = 185$$ Plug $a$ and $b$ into first equation: $$100(-2) + 10(185) + c = 1650$$ $$-200 + 1850 + c = 1650$$ $$c = 1650 - 1650 = 0$$ So, $$R(x) = -2x^2 + 185x$$ 4. **Step 3: Find the cost function $C(x)$** Marginal cost $MC(x)$ is derivative of cost $C(x)$: $$MC(x) = C'(x)$$ Fixed cost = 500, so $$C(x) = \int MC(x) dx + 500$$ Since $MC(x)$ is not explicitly given, assume it is constant or given by the problem (not provided here). For this problem, we focus on profit maximization using revenue and cost. 5. **Step 4: Define profit function $P(x)$** $$P(x) = R(x) - C(x)$$ 6. **Step 5: Maximize profit** Profit is maximized where marginal profit is zero: $$P'(x) = R'(x) - C'(x) = 0$$ Given $R(x) = -2x^2 + 185x$, then $$R'(x) = -4x + 185$$ Assuming marginal cost $MC(x)$ is constant or given, set $$R'(x) = MC(x)$$ Since marginal cost is not given explicitly, but fixed cost is 500, and marginal cost is usually constant or given, we assume marginal cost is constant or zero for this problem. Set $$-4x + 185 = MC$$ If marginal cost is constant, say $MC = m$, then $$x = \frac{185 - m}{4}$$ From the options given (42, 38, 27, 53), test which $x$ satisfies the profit maximization condition. 7. **Step 6: Check options** Calculate $R'(x)$ for each option: - For $x=42$: $R'(42) = -4(42) + 185 = -168 + 185 = 17$ - For $x=38$: $R'(38) = -152 + 185 = 33$ - For $x=27$: $R'(27) = -108 + 185 = 77$ - For $x=53$: $R'(53) = -212 + 185 = -27$ Marginal cost should be positive and close to these values. The only positive and reasonable marginal cost is at $x=42$ with $R'(42) = 17$. **Therefore, the number of laptops to sell to maximize profit is 42.**