1. **State the problem:** We have the temperature function along a metal rod: $$T(x) = Jx - 5x^2 - 0.5x^3$$ with $J=62$ and $x$ in $[0,10]$. We need to: - Find stationary points by solving $T'(x)=0$. - Classify each stationary point using $T''(x)$. - Check endpoints $x=0$ and $x=10$ to find the global maximum temperature. 2. **Find the first derivative $T'(x)$:** $$T'(x) = \frac{d}{dx}(62x - 5x^2 - 0.5x^3) = 62 - 10x - 1.5x^2$$ 3. **Find stationary points by solving $T'(x)=0$:** $$62 - 10x - 1.5x^2 = 0$$ Rewrite as: $$-1.5x^2 - 10x + 62 = 0$$ Multiply both sides by $-1$ for simplicity: $$1.5x^2 + 10x - 62 = 0$$ Use quadratic formula: $$x = \frac{-10 \pm \sqrt{10^2 - 4 \times 1.5 \times (-62)}}{2 \times 1.5}$$ Calculate discriminant: $$\Delta = 100 + 372 = 472$$ Square root: $$\sqrt{472} \approx 21.725$$ Calculate roots: $$x_1 = \frac{-10 + 21.725}{3} = \frac{11.725}{3} \approx 3.908$$ $$x_2 = \frac{-10 - 21.725}{3} = \frac{-31.725}{3} \approx -10.575$$ Since $x$ is in $[0,10]$, discard $x_2$. 4. **Find the second derivative $T''(x)$:** $$T''(x) = \frac{d}{dx}T'(x) = \frac{d}{dx}(62 - 10x - 1.5x^2) = -10 - 3x$$ 5. **Classify the stationary point $x \approx 3.908$:** Calculate $T''(3.908)$: $$T''(3.908) = -10 - 3(3.908) = -10 - 11.724 = -21.724$$ Since $T''(3.908) < 0$, the point is a local maximum. 6. **Check endpoints $x=0$ and $x=10$:** Calculate $T(0)$: $$T(0) = 62 \times 0 - 5 \times 0^2 - 0.5 \times 0^3 = 0$$ Calculate $T(10)$: $$T(10) = 62 \times 10 - 5 \times 10^2 - 0.5 \times 10^3 = 620 - 500 - 500 = -380$$ 7. **Calculate temperature at stationary point $x \approx 3.908$:** $$T(3.908) = 62(3.908) - 5(3.908)^2 - 0.5(3.908)^3$$ Calculate each term: $$62 \times 3.908 \approx 242.296$$ $$5 \times (3.908)^2 = 5 \times 15.27 = 76.35$$ $$0.5 \times (3.908)^3 = 0.5 \times 59.66 = 29.83$$ Sum: $$242.296 - 76.35 - 29.83 = 136.116$$ 8. **Determine global maximum:** Compare values: - $T(0) = 0$ - $T(10) = -380$ - $T(3.908) \approx 136.116$ The global maximum temperature is approximately $136.12$ °C at $x \approx 3.91$ meters.