What is the product of (2p+7)(3p^2 +4p-3)
1 answer:
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Queremos maximizar el precio de tal forma que los ingresos no disminuyan.
Ese maximo precio es: $14,040.6
Sabemos que actualmente el precio es:
p = $6,000
El número de clientes es:
C = 120
Actualmente los ingresos son el producto de esos dos números, es decir:
ingresos = $6,000*120 = $720,000
Ahora sabemos que por cada incremento de $700 en el precio, el número de clientes decrece en 10.
Entonces podemos escribir el número de clientes como una ecuación lineal.
C(p) = a*p + b
tal que tenemos dos puntos en esa linea:
($6,000, 120)
($6,700, 110)
La pendiente es:
Entonces tenemos:
C(p) = (-10/$700)*p + b
Sabemos que:
C($6,000) = 120 = (-10/$700)*$6,000 + b
120 = -85.71 + b
120 + 85.71 = b =
Entonces la ecuación lineal es:
C(p) = (-10/$700)*p + 205.71
Los ingresos serán dados por:
ingresos = C(p)*p = (-10/$700)*p^2 + 205.71*p
Y queremos maximizar p de tal forma que esto sea igual a lo que obtuvimos antes:
(-10/$700)*p^2 + 205.71*p = $720,000
Entonces debemos resolver la ecuación cuadratica:
(-10/$700)*p^2 + 205.71*p - $720,000 = 0.
Las soluciones son dadas por la formula de Bhaskara.
La solución de maximo valor es:
p = (-205.71 - 195.45)/(-20/$700) = $14,040.6
Sí quieres aprender más, puedes leer.
Answer:
a) 0.1587 (or 15.87%)
b) 0.0478 (or 4.78%)
c) 0.7935 (or 79.35%)
Step-by-step explanation:
Find the following probabilities:
(a) the thickness is less than 3.0 mm
We need to find the area under the Normal curve with a mean of 4.5 mm and a standard deviation of 1.5 mm to the left of 3 mm
<h3>(See picture 1 attached) </h3>We can do that with the help of a calculator or a spreadsheet.
<em>In Excel use </em>
<em>NORMDIST(3,4.5,1.5,1) </em>
<em>In OpenOffice Calc use </em>
<em>NORMDIST(3;4.5;1.5;1) </em>
and we get the value 0.1587 (or 15.87%)
(b) the thickness is more than 7.0 mm
Now we need the area to the right of 7 (1 - the area to the left of 7)
<h3>(See picture 2 attached) </h3><em>In Excel use </em>
<em>1-NORMDIST(7,4.5,1.5,1) </em>
<em>In OpenOffice Calc use </em>
<em>1-NORMDIST(7;4.5;1.5;1) </em>
and we get the value 0.0478 (or 4.78%)
(c) the thickness is between 3.0 mm and 7.0 mm
We are looking for the area between 3 and 7
<h3>(See Picture 3) </h3>Since the area under the Normal equals 1, we have
Area to the left of 3 + Area between 3 and 7 + Area to the right of 7 = 1
Hence,
0.1587 + Area between 3 and 7 + 0.0478 = 1
and
Area between 3 and 7 = 1 - 0.1587 - 0.0478 = 0.7935 (or 79.35%)
