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Se ha observado que en una carretera de salida de una gran ciudad la velocidad de los coches entre las 2 horas y las 6 horas de la tarde viene dada por: V=t^(3)-15t^(2)+72 t+8 para t in[2,6] a. ¿A qué hora circulan los coches con mayor velocidad? b. ¿A qué hora circulan los coches con menor velocidad?

55. Se ha observado que en una carretera de salida de una gran ciudad la velocidad de los coches entre las 22 horas y las 66 horas de la tarde viene dada por: V=t3−15t2+72t+8 V=t^{3}-15 t^{2}+72 t+8 para t∈[2,6] t \in[2,6] \newlinea. ¿A qué hora circulan los coches con mayor velocidad?\newlineb. ¿A qué hora circulan los coches con menor velocidad?

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Q. 55. Se ha observado que en una carretera de salida de una gran ciudad la velocidad de los coches entre las 22 horas y las 66 horas de la tarde viene dada por: V=t3−15t2+72t+8 V=t^{3}-15 t^{2}+72 t+8 para t∈[2,6] t \in[2,6] \newlinea. ¿A qué hora circulan los coches con mayor velocidad?\newlineb. ¿A qué hora circulan los coches con menor velocidad?
  1. Calculate Derivative: Calculate the derivative of V(t)V(t) to find the critical points where the speed could be maximum or minimum.\newlineV′(t)=3t2−30t+72V'(t) = 3t^2 - 30t + 72.
  2. Find Critical Points: Set the derivative equal to zero to find critical points. \newline3t2−30t+72=03t^2 - 30t + 72 = 0.\newlineDivide through by 33: t2−10t+24=0t^2 - 10t + 24 = 0.
  3. Factorize Quadratic Equation: Factorize the quadratic equation.\newline(t−6)(t−4)=0(t - 6)(t - 4) = 0.\newlineSo, t=6t = 6 or t=4t = 4.
  4. Evaluate Speed at Points: Evaluate V(t)V(t) at t=2,4,6t = 2, 4, 6 to determine which gives the maximum and minimum speeds.V(2)=23−15⋅22+72⋅2+8=8−60+144+8=100V(2) = 2^3 - 15 \cdot 2^2 + 72 \cdot 2 + 8 = 8 - 60 + 144 + 8 = 100.V(4)=43−15⋅42+72⋅4+8=64−240+288+8=120V(4) = 4^3 - 15 \cdot 4^2 + 72 \cdot 4 + 8 = 64 - 240 + 288 + 8 = 120.V(6)=63−15⋅62+72⋅6+8=216−540+432+8=116V(6) = 6^3 - 15 \cdot 6^2 + 72 \cdot 6 + 8 = 216 - 540 + 432 + 8 = 116.

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