1. **Problem statement:** Given the demand function $P = 190 - 0.6x$ and total cost function $TC = 40 + 30x + 0.4x^2$, where $P$ is price per unit and $x$ is units produced and sold, find: (i) Revenue function (ii) Profit function (iii) Number of units to maximize profit (iv) Maximum profit 2. **Revenue function:** Revenue $R$ is price times quantity sold: $$R = P \times x = (190 - 0.6x) x = 190x - 0.6x^2$$ 3. **Profit function:** Profit $\pi$ is revenue minus total cost: $$\pi = R - TC = (190x - 0.6x^2) - (40 + 30x + 0.4x^2)$$ Simplify: $$\pi = 190x - 0.6x^2 - 40 - 30x - 0.4x^2 = (190x - 30x) - (0.6x^2 + 0.4x^2) - 40 = 160x - x^2 - 40$$ 4. **Maximize profit:** To find $x$ that maximizes profit, take derivative and set to zero: $$\frac{d\pi}{dx} = 160 - 2x = 0$$ Solve for $x$: $$2x = 160$$ $$x = \cancel{\frac{2x}{2}}^{\cancel{2}} = \cancel{\frac{160}{2}}^{\cancel{2}} = 80$$ 5. **Maximum profit:** Substitute $x=80$ into profit function: $$\pi = 160(80) - (80)^2 - 40 = 12800 - 6400 - 40 = 6359$$ 6. **Summary for part (a):** (i) Revenue function: $R = 190x - 0.6x^2$ (ii) Profit function: $\pi = 160x - x^2 - 40$ (iii) Units to maximize profit: $x = 80$ (iv) Maximum profit: $6359$ --- 7. **Part (b): Solve quadratic equation $x^2 - 7x + 10 = 0$** Factor: $$(x - 5)(x - 2) = 0$$ Solutions: $$x = 5 \quad \text{or} \quad x = 2$$ --- 8. **Part (c): Cost of machines** Let cost of drilling machine be $d$, then lathe machine cost is $5d$. Given: $$2(5d) + 5d = 7500$$ Simplify: $$10d + 5d = 7500$$ $$15d = 7500$$ $$d = \cancel{\frac{15d}{15}}^{\cancel{15}} = \cancel{\frac{7500}{15}}^{\cancel{15}} = 500$$ Lathe machine cost: $$5d = 5 \times 500 = 2500$$ --- 9. **Part (d): Solve for $q$ in** $$\frac{q}{2} + \frac{q}{3} + 3 = 2 + \frac{q}{6}$$ Multiply both sides by 6 to clear denominators: $$6 \times \left(\frac{q}{2} + \frac{q}{3} + 3\right) = 6 \times \left(2 + \frac{q}{6}\right)$$ $$3q + 2q + 18 = 12 + q$$ Combine like terms: $$5q + 18 = 12 + q$$ Subtract $q$ from both sides: $$5q - q + 18 = 12$$ $$4q + 18 = 12$$ Subtract 18: $$4q = 12 - 18 = -6$$ Divide by 4: $$q = \cancel{\frac{4q}{4}}^{\cancel{4}} = \cancel{\frac{-6}{4}}^{\cancel{2}} = -\frac{3}{2} = -1.5$$