1. **Problem 1: Find the dimensions of a rectangle with area 1000 m² that minimize the perimeter.** 2. Given: - Area $A = xy = 1000$ - Perimeter $P = 2x + 2y$ 3. Express $y$ in terms of $x$ using the area constraint: $$y = \frac{1000}{x}$$ 4. Substitute into the perimeter formula: $$P(x) = 2x + 2\frac{1000}{x} = 2x + \frac{2000}{x}$$ 5. To minimize $P(x)$, find the derivative and set it to zero: $$P'(x) = 2 - \frac{2000}{x^2} = 0$$ 6. Solve for $x$: $$2 = \frac{2000}{x^2} \implies 2x^2 = 2000 \implies x^2 = 1000 \implies x = \sqrt{1000} = 10\sqrt{10} \approx 31.62$$ 7. Find $y$: $$y = \frac{1000}{31.62} \approx 31.62$$ 8. So, the rectangle is a square with dimensions approximately $31.62, 31.62$ meters. --- 1. **Problem 2: Find the point on the line $y = 5x + 3$ closest to the origin.** 2. The distance squared from origin to point $(x,y)$ is: $$D^2 = x^2 + y^2$$ 3. Substitute $y = 5x + 3$: $$D^2 = x^2 + (5x + 3)^2 = x^2 + 25x^2 + 30x + 9 = 26x^2 + 30x + 9$$ 4. Minimize $D^2$ by taking derivative and setting to zero: $$\frac{d}{dx}D^2 = 52x + 30 = 0 \implies x = -\frac{30}{52} = -\frac{15}{26}$$ 5. Find $y$: $$y = 5\left(-\frac{15}{26}\right) + 3 = -\frac{75}{26} + \frac{78}{26} = \frac{3}{26}$$ 6. The closest point is $\left(-\frac{15}{26}, \frac{3}{26}\right)$. --- **Final answers:** - Rectangle dimensions: $31.62, 31.62$ - Closest point on line: $\left(-\frac{15}{26}, \frac{3}{26}\right)$