1. Problem 07a: Find the slope of the parabola $y=\sqrt{x}$ and write the equation of the tangent line at $x=4$. 2. To find the slope of the tangent line, we need the derivative $\frac{dy}{dx}$ of $y=\sqrt{x} = x^{1/2}$. 3. Using the power rule, $\frac{dy}{dx} = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$. 4. Evaluate the slope at $x=4$: $m = \frac{1}{2\sqrt{4}} = \frac{1}{2 \times 2} = \frac{1}{4}$. 5. Find the point on the curve at $x=4$: $y = \sqrt{4} = 2$. 6. Equation of the tangent line using point-slope form: $y - y_1 = m(x - x_1)$, where $(x_1,y_1) = (4,2)$. 7. Substitute values: $y - 2 = \frac{1}{4}(x - 4)$. 8. Simplify: $y = \frac{1}{4}x - 1 + 2 = \frac{1}{4}x + 1$. --- 9. Problem 07b: Find $\frac{d^2y}{dx^2}$ for $y = \sqrt[3]{1 + x^2} = (1 + x^2)^{1/3}$. 10. First derivative using chain rule: $$\frac{dy}{dx} = \frac{1}{3}(1 + x^2)^{-2/3} \times 2x = \frac{2x}{3(1 + x^2)^{2/3}}$$ 11. Second derivative using quotient and chain rules: $$\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{2x}{3(1 + x^2)^{2/3}} \right)$$ 12. Let $u=2x$, $v=3(1 + x^2)^{2/3}$, then $$\frac{d^2y}{dx^2} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ 13. Compute derivatives: $$\frac{du}{dx} = 2$$ $$\frac{dv}{dx} = 3 \times \frac{2}{3} (1 + x^2)^{-1/3} \times 2x = 2 \times 2x (1 + x^2)^{-1/3} = 4x (1 + x^2)^{-1/3}$$ 14. Substitute: $$\frac{d^2y}{dx^2} = \frac{3(1 + x^2)^{2/3} \times 2 - 2x \times 4x (1 + x^2)^{-1/3}}{9(1 + x^2)^{4/3}}$$ 15. Simplify numerator: $$6(1 + x^2)^{2/3} - 8x^2 (1 + x^2)^{-1/3}$$ 16. Factor out $(1 + x^2)^{-1/3}$: $$ (1 + x^2)^{-1/3} \left[ 6(1 + x^2) - 8x^2 \right] = (1 + x^2)^{-1/3} (6 + 6x^2 - 8x^2) = (1 + x^2)^{-1/3} (6 - 2x^2)$$ 17. Denominator is $9(1 + x^2)^{4/3}$, so $$\frac{d^2y}{dx^2} = \frac{(1 + x^2)^{-1/3} (6 - 2x^2)}{9(1 + x^2)^{4/3}} = \frac{6 - 2x^2}{9(1 + x^2)^{5/3}}$$ --- 18. Problem 08a: Minimize cost function $C = \frac{51200}{q} + 80q + 750000$ with respect to $q$. 19. Take derivative: $$\frac{dC}{dq} = -\frac{51200}{q^2} + 80$$ 20. Set derivative to zero for minimum: $$-\frac{51200}{q^2} + 80 = 0 \Rightarrow 80 = \frac{51200}{q^2} \Rightarrow q^2 = \frac{51200}{80} = 640$$ 21. Solve for $q$: $$q = \sqrt{640} = 8\sqrt{10} \approx 25.298$$ --- 22. Problem 08b: Find minimum cost by substituting $q = 8\sqrt{10}$ into $C$. 23. Compute: $$C = \frac{51200}{8\sqrt{10}} + 80 \times 8\sqrt{10} + 750000$$ 24. Simplify terms: $$\frac{51200}{8\sqrt{10}} = \frac{51200}{8} \times \frac{1}{\sqrt{10}} = 6400 \times \frac{1}{\sqrt{10}} = 6400 \times \frac{\sqrt{10}}{10} = 640 \sqrt{10}$$ $$80 \times 8 \sqrt{10} = 640 \sqrt{10}$$ 25. Sum terms: $$C = 640 \sqrt{10} + 640 \sqrt{10} + 750000 = 1280 \sqrt{10} + 750000$$ 26. Approximate $\sqrt{10} \approx 3.1623$: $$C \approx 1280 \times 3.1623 + 750000 = 4047 + 750000 = 754047$$ Final answers: - 07a) Tangent line at $x=4$ is $y = \frac{1}{4}x + 1$. - 07b) Second derivative is $\frac{d^2y}{dx^2} = \frac{6 - 2x^2}{9(1 + x^2)^{5/3}}$. - 08a) Optimal order size $q = 8\sqrt{10} \approx 25.298$. - 08b) Minimum cost $C \approx 754047$.