1. **Find the point on the line $y = 5x + 3$ closest to the origin.** 2. The distance $d$ from the origin $(0,0)$ to a point $(x,y)$ is given by the distance formula: $$d = \sqrt{x^2 + y^2}$$ 3. Since $y = 5x + 3$, substitute $y$ into the distance formula: $$d = \sqrt{x^2 + (5x + 3)^2}$$ 4. To minimize $d$, minimize $d^2$ (to avoid the square root): $$d^2 = x^2 + (5x + 3)^2 = x^2 + 25x^2 + 30x + 9 = 26x^2 + 30x + 9$$ 5. Find the derivative of $d^2$ with respect to $x$ and set it to zero to find critical points: $$\frac{d}{dx} d^2 = 52x + 30 = 0$$ 6. Solve for $x$: $$52x = -30 \implies x = -\frac{30}{52} = -\frac{15}{26}$$ 7. Find $y$ using $y = 5x + 3$: $$y = 5 \times -\frac{15}{26} + 3 = -\frac{75}{26} + 3 = -\frac{75}{26} + \frac{78}{26} = \frac{3}{26}$$ 8. So, the point closest to the origin is: $$(x,y) = \left(-\frac{15}{26}, \frac{3}{26}\right)$$ --- 9. **Find the dimensions of a box with square base and open top, volume 4000 cm³, minimizing material used.** 10. Let the side of the square base be $x$ cm and the height be $h$ cm. 11. Volume constraint: $$V = x^2 h = 4000$$ 12. Surface area (material used) to minimize (open top): $$A = x^2 + 4xh$$ 13. From volume, express $h$: $$h = \frac{4000}{x^2}$$ 14. Substitute $h$ into $A$: $$A = x^2 + 4x \times \frac{4000}{x^2} = x^2 + \frac{16000}{x}$$ 15. Find derivative of $A$ with respect to $x$: $$\frac{dA}{dx} = 2x - \frac{16000}{x^2}$$ 16. Set derivative to zero to find critical points: $$2x - \frac{16000}{x^2} = 0 \implies 2x = \frac{16000}{x^2} \implies 2x^3 = 16000 \implies x^3 = 8000$$ 17. Solve for $x$: $$x = \sqrt[3]{8000} = 20$$ 18. Find $h$: $$h = \frac{4000}{20^2} = \frac{4000}{400} = 10$$ 19. Dimensions that minimize material: - Side of base: 20 cm - Height: 10 cm