1. **State the problem:** Find the derivative $h'(x)$ of the function $$h(x) = \frac{1}{60}x^2 - \frac{1}{4}x + \frac{3}{5}$$ Then use $h'(x)$ to find the horizontal distance $x$ where the swimmer reaches the greatest depth (maximum or minimum point). Finally, find the greatest depth by evaluating $h(x)$ at that $x$. 2. **Find the derivative $h'(x)$:** Use the power rule: $$\frac{d}{dx} x^n = n x^{n-1}$$ $$h'(x) = \frac{d}{dx} \left( \frac{1}{60}x^2 - \frac{1}{4}x + \frac{3}{5} \right) = \frac{1}{60} \cdot 2x - \frac{1}{4} \cdot 1 + 0 = \frac{2}{60}x - \frac{1}{4} = \frac{1}{30}x - \frac{1}{4}$$ 3. **Find the horizontal distance $x$ where $h'(x) = 0$ (critical point):** Set derivative equal to zero: $$\frac{1}{30}x - \frac{1}{4} = 0$$ Add $\frac{1}{4}$ to both sides: $$\frac{1}{30}x = \frac{1}{4}$$ Multiply both sides by 30: $$x = 30 \times \frac{1}{4}$$ Show cancellation: $$x = \cancel{30} \times \frac{1}{4} = 7.5$$ So, the swimmer reaches greatest depth at $x = 7.5$ meters. 4. **Find the greatest depth $h(7.5)$:** Substitute $x=7.5$ into $h(x)$: $$h(7.5) = \frac{1}{60} (7.5)^2 - \frac{1}{4} (7.5) + \frac{3}{5}$$ Calculate each term: $$\frac{1}{60} (7.5)^2 = \frac{1}{60} \times 56.25 = 0.9375$$ $$- \frac{1}{4} (7.5) = -1.875$$ $$\frac{3}{5} = 0.6$$ Sum all: $$h(7.5) = 0.9375 - 1.875 + 0.6 = -0.3375$$ **Final answer:** - Derivative: $$h'(x) = \frac{1}{30}x - \frac{1}{4}$$ - Horizontal distance for greatest depth: $$x = 7.5$$ meters - Greatest depth: $$h(7.5) = -0.3375$$ meters