1. Problem (a): Write an expression for $y$ in terms of $x$. Since the problem context is missing, assume $y$ is a function of $x$ related to the garden and path. Without additional info, we cannot write $y$ explicitly here. 2. Problem (b): Show total area $A = 96 + 4x + \frac{320}{x}$. Assuming the garden area is fixed and the path width is $x$, the total area includes garden plus path. Let garden area be $96$. The path adds area along the sides, giving terms $4x$ and $\frac{320}{x}$ from length and width adjustments. 3. Problem (c): Find minimum area of the path. Given $A = 96 + 4x + \frac{320}{x}$, minimize $A$ with respect to $x > 0$. Step 1: Differentiate $A$ with respect to $x$: $$\frac{dA}{dx} = 4 - \frac{320}{x^2}$$ Step 2: Set derivative to zero for critical points: $$4 - \frac{320}{x^2} = 0 \implies 4 = \frac{320}{x^2} \implies x^2 = \frac{320}{4} = 80$$ Step 3: Solve for $x$: $$x = \sqrt{80} = 4\sqrt{5}$$ Step 4: Find minimum area by substituting $x$ back into $A$: $$A = 96 + 4(4\sqrt{5}) + \frac{320}{4\sqrt{5}} = 96 + 16\sqrt{5} + \frac{320}{4\sqrt{5}}$$ Simplify last term: $$\frac{320}{4\sqrt{5}} = \frac{80}{\sqrt{5}} = 80 \times \frac{\sqrt{5}}{5} = 16\sqrt{5}$$ So, $$A = 96 + 16\sqrt{5} + 16\sqrt{5} = 96 + 32\sqrt{5}$$ 4. Problem 7(a): Height of the tower is $h(0)$. $$h(0) = -2(0)^2 + 20(0) + 8 = 8$$ 5. Problem 7(b): Height at $t=5$ seconds. $$h(5) = -2(5)^2 + 20(5) + 8 = -2(25) + 100 + 8 = -50 + 108 = 58$$ 6. Problem 7(c): Find $n$ when $h(n) = 0$. Solve $$-2n^2 + 20n + 8 = 0$$ Divide by -2: $$n^2 - 10n - 4 = 0$$ Use quadratic formula: $$n = \frac{10 \pm \sqrt{(-10)^2 - 4(1)(-4)}}{2} = \frac{10 \pm \sqrt{100 + 16}}{2} = \frac{10 \pm \sqrt{116}}{2}$$ $$\sqrt{116} = 2\sqrt{29}$$ So, $$n = \frac{10 \pm 2\sqrt{29}}{2} = 5 \pm \sqrt{29}$$ Since $t \geq 0$, take positive root: $$n = 5 + \sqrt{29} \approx 5 + 5.385 = 10.385$$ 7. Problem 7(d): Find $h'(t)$. $$h'(t) = \frac{d}{dt}(-2t^2 + 20t + 8) = -4t + 20$$ 8. Problem 7(e): Maximum height and time. Set $h'(t) = 0$: $$-4t + 20 = 0 \implies t = 5$$ Calculate max height: $$h(5) = 58$$ (from step 5) 9. Problem 7(f): Total time above tower height (8 m). Solve $h(t) = 8$: $$-2t^2 + 20t + 8 = 8 \implies -2t^2 + 20t = 0 \implies -2t(t - 10) = 0$$ Roots: $t=0$ and $t=10$. The cannon-ball is above 8 m between $t=0$ and $t=10$. Total time above tower height: $$10 - 0 = 10 \text{ seconds}$$