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Vlad
10 days ago
11

Could you help me to solve the problem below the cost for producing x items is 50x+300 and the revenue for selling x items is 90

x−0.5^2 the company makes is how much it takes in (revenue) minus how much it spends (cost). Recall that profit is revenue minus cost. Set up an expression for the profit from producing and selling x items. Find two values of x that will create a profit of $50. Is it possible for the company to make a profit of $2,500? Please explain
Mathematics
1 answer:
tester [12.3K]10 days ago
5 0

Answer:

Función de beneficios: P(x) = -0.5x^2 + 40x - 300

beneficio de $50: x = 10 y x = 70

NO es posible generar un ingreso de $2,500, ya que el máximo beneficio es de $500

Step-by-step explanation:

(Suponiendo que la función de ingresos es 90x−0.5x^2)

La función de costo está dada por:

C(x) = 50x + 300

Y la función de ingresos se expresa como:

R(x) = 90x - 0.5x^2

La función de beneficios se determina restando los costos de los ingresos, por lo que tenemos:

P(x) = R(x) - C(x)

P(x) = 90x - 0.5x^2 - 50x - 300

P(x) = -0.5x^2 + 40x - 300

Para encontrar los puntos donde los beneficios son $50, usamos P(x) = 50 y luego hallamos los valores de x:

50 = -0.5x^2 + 40x - 300

-0.5x^2 + 40x - 350 = 0

x^2 - 80x + 700 = 0

Aplicando la fórmula de Bhaskara, obtenemos:

\Delta = b^2 - 4ac = (-80)^2 - 4*700 = 3600

x_1 = (-b + \sqrt{\Delta})/2a = (80 + 60)/2 = 70

x_2 = (-b - \sqrt{\Delta})/2a = (80 - 60)/2 = 10

Los valores de x que resultan en un beneficio de $50 son x = 10 y x = 70

Para verificar si es posible obtener un beneficio de $2,500, necesitamos considerar el beneficio máximo, que es el máximo de la función P(x).

El valor máximo de P(x) se encuentra en el vértice. La coordenada x del vértice se calcula como:

x_v = -b/2a = 80/2 = 40

Con este valor de x, podemos calcular el beneficio máximo:

P(40) = -0.5(40)^2 + 40*40 - 300 = $500

El beneficio máximo es de $500, así que NO es posible generar un beneficio de $2,500.

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Answer:

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Step-by-step explanation:

The range indicates the values lie between the minimum and maximum figures.

Given that the values may vary by 1.4%, the range can be calculated as:

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