Función de beneficios = P(x) =-0.01x² + 3x - 100
Ganancias máximos = 125
Y ocurre cuando x = 150 unidades
Profits Function = P(x) = -0.01x² + 3x - 100
Maximum profits = 125
And it occurs when x = 150 units
Step-by-step explanation:
Ganancias = Ingresos - Costos
Ingresos = R (x) = 5x - 0.01x²
Costos = C (x) = 100 + 2x
Función de ganancias = P (x) = R (x) - C (x)
= (5x - 0.01x²) - (100 + 2x)
= 5x - 0.01x² - 100 - 2x
= 3x - 0.01x² - 100
P (x) = -0.01x² + 3x - 100
Para obtener las máximas ganancias, utilizamos el análisis de diferenciación para la función de ganancias.
En el nivel de Ganancias máximo, (dP / dx) = 0 y (d²P / dx²) < 0
P(x) = -0.01x² + 3x - 100
(dP/dx) = -0.02x + 3
Al máximo Ganancias, (dP/dx) = 0
(dP/dx) = -0.02x + 3 = 0
x = (3/0.02) = 150 unidades
Este es el nivel de producto que corresponde a las ganancias máximas.
Para verificar si este es realmente el punto máximo de la función de ganancias,
(dP/dx) = -0.02x + 3
(d²P/dx²) = -0.02 < 0 (lo que demuestra que realmente es el punto de máximo ganancias).
Por lo tanto, la ganancia máxima ocurre cuando x = 150 unidades
P(x) = -0.01x² + 3x - 100
P(x) = -0.01(150²) + 3(150) - 100 = 125
¡¡¡Espero que esto ayude!!!
English Translation
The profits of a business can be determined by subtracting the costs from the income. Suppose that the revenue of a business is represented by the function:
R(x) = 5x - 0.01x²
and the manufacturing costs of the product are represented by
C(x) = 100 + 2x
where x is the number of units of the product. Determine the profits function. Find a function P (x) that represents the profits of the firm and determine the maximum profit.
Solution
Profits = Revenue - Costs
Revenue = R(x) = 5x - 0.01x²
Costs = C(x) = 100 + 2x
Profits function = P(x) = R(x) - C(x)
= (5x - 0.01x²) - (100 + 2x)
= 5x - 0.01x² - 100 - 2x
= 3x - 0.01x² - 100
P(x) = -0.01x² + 3x - 100
To obtain maximum profits, we use differentiation analysis for the profit function.
At the maximum profit level, (dP/dx) = 0 and (d²P/dx²) < 0
P(x) = -0.01x² + 3x - 100
(dP/dx) = -0.02x + 3
At maximum profit, (dP/dx) = 0
(dP/dx) = -0.02x + 3 = 0
x = (3/0.02) = 150 units
This is the product level that corresponds to maximum profits.
To check if this is truly the maximum point of the profit function,
(dP/dx) = -0.02x + 3
(d²P/dx²) = -0.02 < 0 (which shows that it truly is the maximum profit point.
Hence, maximum profit occurs when x = 150 units
P(x) = -0.01x² + 3x - 100
P(x) = -0.01(150²) + 3(150) - 100 = 125
Hope this Helps!!!
