ValdesValles JudithdelCarmen M18S4 Enuntiempo

Proyecto integrador En un tiempo... Nombre: Judith del Carmen Valdés Valles Facilitador: Rogelio Reyes Medina Modulo 18

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Una asociación contra el cáncer de niños se encarga de recolectar tapas desechables con el propósito de venderlas y así obtener una cantidad de dinero extra para continuar con su labor. Según su estadística, la ecuación que representa rep resenta el número de tapas a recolectar es la siguiente f(x)= -x 2 + 12x donde f(x) señala la cantidad de tapas recolectadas y "x" representa el tiempo en semanas. Ligado a esto, la asociación ya cuenta con 30,000 tapas que ha recolectado por su cuenta.

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 Ahora, la pendiente será: m = (65750-65000)/(6.5-5) m = 500 Por tanto, la recta secante será: y-65000 = 500(x-5) y = 500x + 62500 → Recta secante

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Solución y desarrollo: f(x) = -x² + 12 Tapas iniciales 30000 Por tanto, la ecuación que representaría esto es la siguiente: f(x) = (-x² + 12) + 30000 1- f(x) = -1000x² -1000x² + 12000x + 30000 30000 Para buscar la cantidad de tapas máximas debemos derivar e igualar a cero, y buscar el punto máximo: f(x) = -1000x² + 12000x + 30000 f'(x) = -2000x + 12000 = 0 x = 6 → Semana 6  Ahora buscamos buscamos la cantidad de de tapas en la semana semana 6, tenemos: f(6) = -1000(6)² + 12000(6) + 30000

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c) ¿Cuál es la relación que existe entre el tiempo y el número de tapas que se juntaron? Que las primeras semanas se juntaron el mayor numero de tapas, hasta llegar a la semana 6 que es donde llego a su punto mas ato de recolección y después de esa semana el numero de tapas recolectado fue disminuyendo.

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 Ahora buscamos la recta secante, para ello buscamos la pendiente de la misma, tenemos que: x₁ = 5 ∴ f(5) = -1000(5)² + 12000(5) + 30000 = 65000 x₂ = 6.5 ∴ f(6.5) = -1000(6.5)² + 12000(6.55) + 30000 = 65750Ahora, la pendiente

 Ahora, la recta tangente en el punto más alto será: f'(x) = 2000x + 12000El punto más alto será (6,66000), evaluamos y tenemos: f'(6) = -2000(6) + 120000 = 0 ¿Tiene sentido que la pendiente sea cero? Si tiene

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