1. Problem 1: Find the dimensions of a rectangle with perimeter 84 m that maximize the area. 2. The perimeter $P$ of a rectangle with length $l$ and width $w$ is given by: $$P = 2(l + w)$$ Given $P = 84$, we have: $$2(l + w) = 84 \implies l + w = 42$$ 3. The area $A$ of the rectangle is: $$A = l \times w$$ Using $w = 42 - l$, substitute into the area formula: $$A = l(42 - l) = 42l - l^2$$ 4. To maximize the area, take the derivative of $A$ with respect to $l$ and set it to zero: $$\frac{dA}{dl} = 42 - 2l = 0 \implies 2l = 42 \implies l = 21$$ 5. Substitute $l = 21$ back to find $w$: $$w = 42 - 21 = 21$$ 6. So, the rectangle with maximum area given perimeter 84 m is a square with dimensions: $$21, 21$$ meters. 7. Problem 2: Find the dimensions of a rectangle with area 1000 m$^2$ that minimize the perimeter. 8. The area is given by: $$A = l \times w = 1000$$ Express $w$ in terms of $l$: $$w = \frac{1000}{l}$$ 9. The perimeter is: $$P = 2(l + w) = 2\left(l + \frac{1000}{l}\right)$$ 10. To minimize $P$, take the derivative with respect to $l$ and set it to zero: $$\frac{dP}{dl} = 2\left(1 - \frac{1000}{l^2}\right) = 0 \implies 1 = \frac{1000}{l^2} \implies l^2 = 1000 \implies l = \sqrt{1000}$$ 11. Calculate $l$: $$l = \sqrt{1000} = 10\sqrt{10} \approx 31.62$$ 12. Find $w$: $$w = \frac{1000}{31.62} \approx 31.62$$ 13. The rectangle with minimum perimeter for area 1000 m$^2$ is also a square with dimensions: $$31.62, 31.62$$ meters. Final answers: 1. Dimensions: 21, 21 2. Dimensions: 31.62, 31.62