Los ingresos mensuales de un fabricante de zapatos están dados por la función l(z)=1000z-2z^2, donde z es la cantidad de pares de zapatos que fabrica en el mes.
a) Qué cantidad de pares debe fabricar mensualmente para obtener el mayor ingreso?
b) ¿Cuales son los ingresos si se fabrican 125 pares de zapatos? ¿y 375 pares?
c) ¿A partir de que cantidad de pares comienza a tener pérdidas?

a) 250 pairs

b) 125 pairs -> profit = 93750

375 pairs -> profit = 93750

c) 500 pairs

Step-by-step explanation:

a)

To find the value of z that gives maximum profit, we just need to find the vertix of the quadratic function, which can be found with the formula:

z_vertix = -b / 2a

Where for this case a = -2 and b = 1000

So we have:

z_vertix = -1000/(-4) = 250

The maximum profit is given with 250 pairs of shoes.

b)

Using the values z = 125 and z = 375, we have:

l(125) = 1000*125 - 2*125^2 = 93750

l(375) = 1000*375 - 2*375^2 = 93750

c)

First we need to find the roots of the function l(z):

l(z) = 1000z - 2z^2 = z(1000 - 2z)

making l(z) = 0, we have z = 0 or z = 500

As we have a<0, the concavity of the quadratic equation is downwards, so for a number of pairs bigger than 500 we have a negative profit.

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step-by-step explanation:

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