1. **Problem statement:** We have a roller-coaster track modeled by a cubic function $$y = ax^3 + bx^2 + cx + d$$ for $$0 \leq x < 90$$ meters. The track passes through points: - A at $$x=0, y=30$$ (starting point), - C at $$x=50, y=30$$, - D at $$x=80, y=30$$, and has a minimum height of 6 meters somewhere between these points. We need to find the maximum height the roller-coaster reaches. 2. **Key facts and formulas:** - Since $$y$$ is cubic, its derivative $$y' = 3ax^2 + 2bx + c$$ gives critical points (local maxima or minima). - At local extrema, $$y' = 0$$. - The minimum height is 6 meters at some $$x = x_{min}$$. - The maximum height is at another critical point where $$y' = 0$$. 3. **Set up equations from given points:** From points A, C, D: $$ \begin{cases} y(0) = d = 30 \\ y(50) = 125000a + 2500b + 50c + d = 30 \\ y(80) = 512000a + 6400b + 80c + d = 30 \end{cases} $$ 4. **Conditions for extrema:** Let the two critical points be $$x_1$$ and $$x_2$$ with $$x_1 < x_2$$. - At $$x_1$$, the minimum height is 6, so $$y(x_1) = 6$$. - At $$x_2$$, the maximum height is $$y(x_2)$$ (unknown). Since $$y' = 3ax^2 + 2bx + c = 0$$ at both points, the quadratic has roots $$x_1$$ and $$x_2$$. 5. **Express derivative roots:** $$ 3a x^2 + 2b x + c = 3a (x - x_1)(x - x_2) = 3a (x^2 - (x_1 + x_2)x + x_1 x_2) $$ Matching coefficients: $$ 2b = -3a (x_1 + x_2), \quad c = 3a x_1 x_2 $$ 6. **Use minimum height condition:** $$ 6 = y(x_1) = a x_1^3 + b x_1^2 + c x_1 + d $$ Substitute $$b$$ and $$c$$ in terms of $$a, x_1, x_2$$ and $$d=30$$: $$ b = -\frac{3a}{2} (x_1 + x_2), \quad c = 3a x_1 x_2 $$ So: $$ 6 = a x_1^3 + \left(-\frac{3a}{2} (x_1 + x_2)\right) x_1^2 + (3a x_1 x_2) x_1 + 30 $$ Simplify: $$ 6 = a x_1^3 - \frac{3a}{2} x_1^2 (x_1 + x_2) + 3a x_1^2 x_2 + 30 $$ $$ 6 - 30 = a x_1^3 - \frac{3a}{2} x_1^3 - \frac{3a}{2} x_1^2 x_2 + 3a x_1^2 x_2 $$ $$ -24 = a \left(x_1^3 - \frac{3}{2} x_1^3 - \frac{3}{2} x_1^2 x_2 + 3 x_1^2 x_2\right) $$ $$ -24 = a \left(-\frac{1}{2} x_1^3 + \frac{3}{2} x_1^2 x_2\right) = a \frac{1}{2} x_1^2 (3 x_2 - x_1) $$ So: $$ a = \frac{-48}{x_1^2 (3 x_2 - x_1)} $$ 7. **Use points C and D to form two equations:** At $$x=50$$: $$ 30 = a (50)^3 + b (50)^2 + c (50) + 30 $$ Subtract 30 both sides: $$ 0 = 125000 a + 2500 b + 50 c $$ At $$x=80$$: $$ 0 = 512000 a + 6400 b + 80 c $$ Substitute $$b$$ and $$c$$: $$ 0 = 125000 a + 2500 \left(-\frac{3a}{2} (x_1 + x_2)\right) + 50 (3a x_1 x_2) = a \left(125000 - 3750 (x_1 + x_2) + 150 x_1 x_2\right) $$ $$ 0 = 512000 a + 6400 \left(-\frac{3a}{2} (x_1 + x_2)\right) + 80 (3a x_1 x_2) = a \left(512000 - 9600 (x_1 + x_2) + 240 x_1 x_2\right) $$ Since $$a \neq 0$$, set the parentheses to zero: $$ \begin{cases} 125000 - 3750 (x_1 + x_2) + 150 x_1 x_2 = 0 \\ 512000 - 9600 (x_1 + x_2) + 240 x_1 x_2 = 0 \end{cases} $$ 8. **Solve the system for $$x_1 + x_2$$ and $$x_1 x_2$$:** Multiply first equation by 64: $$ 8000000 - 240000 (x_1 + x_2) + 9600 x_1 x_2 = 0 $$ Multiply second equation by 25: $$ 12800000 - 240000 (x_1 + x_2) + 6000 x_1 x_2 = 0 $$ Subtract first from second: $$ 4800000 - 0 + (-3600) x_1 x_2 = 0 \Rightarrow 3600 x_1 x_2 = 4800000 \Rightarrow x_1 x_2 = \frac{4800000}{3600} = 1333.33 $$ Use first equation: $$ 125000 - 3750 (x_1 + x_2) + 150 (1333.33) = 0 $$ $$ 125000 - 3750 (x_1 + x_2) + 200000 = 0 $$ $$ -3750 (x_1 + x_2) = -325000 \Rightarrow x_1 + x_2 = \frac{325000}{3750} = 86.67 $$ 9. **Recall minimum is at $$x_1$$ near 20-30, so $$x_1 < x_2$$ and $$x_1 + x_2 = 86.67$$, $$x_1 x_2 = 1333.33$$. Solve quadratic for $$x$$: $$ x^2 - (x_1 + x_2) x + x_1 x_2 = 0 \Rightarrow x^2 - 86.67 x + 1333.33 = 0 $$ Discriminant: $$ \Delta = 86.67^2 - 4 \times 1333.33 = 7511.11 - 5333.33 = 2177.78 $$ $$ \sqrt{2177.78} \approx 46.67 $$ Roots: $$ \begin{cases} x_1 = \frac{86.67 - 46.67}{2} = 20 \\ x_2 = \frac{86.67 + 46.67}{2} = 66.67 \end{cases} $$ 10. **Calculate $$a$$:** $$ a = \frac{-48}{20^2 (3 \times 66.67 - 20)} = \frac{-48}{400 (200 - 20)} = \frac{-48}{400 \times 180} = \frac{-48}{72000} = -0.0006667 $$ 11. **Calculate $$b$$ and $$c$$:** $$ b = -\frac{3a}{2} (x_1 + x_2) = -\frac{3 \times (-0.0006667)}{2} \times 86.67 = 0.0867 $$ $$ c = 3a x_1 x_2 = 3 \times (-0.0006667) \times 20 \times 66.67 = -2.667 $$ 12. **Find maximum height at $$x_2 = 66.67$$:** $$ y(66.67) = a (66.67)^3 + b (66.67)^2 + c (66.67) + d $$ Calculate each term: $$ (66.67)^3 = 296296.3, \quad (66.67)^2 = 4444.89 $$ $$ y(66.67) = -0.0006667 \times 296296.3 + 0.0867 \times 4444.89 - 2.667 \times 66.67 + 30 $$ $$ y(66.67) = -197.53 + 385.33 - 177.8 + 30 = 40.0 $$ 13. **Final answer:** The maximum height is approximately **40 meters**. Rounded to the nearest ten meters: **40** meters.