Un granjero tiene 100 vacas que pesan 125 kg cada una. El costo diario de mantenimiento de una vaca asciende a $ 5.00. Las vacas ganan peso a razón de 3 kg diarios. El precio en el mercado es ahora de 25 pesos por kg y esta bajando a razón de un centavo diario. ¿cuanto tiempo debe esperar el granjero para obtener una ganancia maxima al vender sus vacas?.

Answer:

25 days

Step-by-step explanation:

Assuming that the farmer waits for n days to get get the maximum profit.

Given that the total number of cows the farmer has, = 100

Currect weight of one cow = 125 kg.

Weight gain rate for one cow = 3 kg/day

So, weight gained by one cow in n days = 3n kg

Therefore, the weight of one cow after n days [tex]= 125 + 3n \;\;kg \cdots(i)[/tex]

Cost for keeping one cow = $5/day.

Cost for keeping one cow for n days = $ 5n

So, Cost for keeping 100 cows for n days [tex]= \$ 5n \times 100 =\$500n \cdots(ii)[/tex]

Current market price = 25 pesos / kg = 125 cents / kg [ as 1 peso = 5 cents]

The falling rate of market price = 1 cent /day.

Fall in price in n days [tex]= 1\times n = n[/tex] cents.

So, market price after n days [tex]= 125-n cent/kg\cdots(iii)[/tex]

By using equations (i) and (iii),

The selling price of one cow after n days [tex]= (125+3n) \times (125-n)[/tex] cents.

So, the selling price of 100 cows after n days [tex]= (125+3n) \times (125-n)\times 100[/tex] cents.

As 1 $ = 100 cents. so

The selling price of 100 cows after n days [tex]=\$ 5(125+3n) (125-n)\cdots(iv)[/tex].

Now, from equations (ii) and (iv)

Net profit, [tex]P = 5(125+3n) (125-n) - 500n\cdots(v)[/tex]

To get the maximum profit, differentiate the profit function in equation (v) with respect to n and equate it to zero to get the value of n, i.e

[tex]\frac {dP}{dn}=0 \\\\\Rightarrow \frac {d}{dn}(5(125+3n) (125-n) - 500n)=0 \\\\\Rightarrow  5[ (125+3n)(-1)+(125-n)(3)]-500=0 \\\\\Rightarrow 5[ -125-3n+375-3n]-500 =0 \\\\\Rightarrow 5[ -125-3n+375-3n]=500\\\\\Rightarrow 250-6n=500/5=100 \\\\\Rightarrow 6n = 250-100=150 \\\\\Rightarrow n= 150/6 = 25.[/tex]

Hence, the farmer has to wait for 25 days to get the maximum profit.

El granjero puede obtener ganancias máximas en el día 229.

En este problema debemos construir una función de ganancia ([tex]G[/tex]), en pesos, igual a la ganancia neta por venta de la carne de vacuno ([tex]N[/tex]) menos los costes de mantenimiento ([tex]M[/tex]). Cada una de las variables ha de considerar el cambio de los indicadores en función del tiempo, entonces:

Función de ganancia

[tex]G = N - M[/tex] (1)

Función de ganancia neta

[tex]N = (125+3\cdot t)\cdot 100\cdot (5-0.01\cdot t)[/tex] (2)

Función de mantenimiento

[tex]M = 5[/tex] (3)

Al aplicar (2) y (3) en (1) tenemos la siguiente fórmula:

[tex]G = (125 + 3\cdot t) \cdot 100 \cdot (5 - 0.01\cdot t) - 5[/tex]

[tex]G = (125 + 3\cdot t) \cdot (500-t) - 5[/tex]

[tex]G = 62500+1500\cdot t -125\cdot t -3\cdot t^{2}-5[/tex]

[tex]G = -3\cdot t^{2}+1375\cdot t +62495[/tex]

El día correspondiente a la máxima ganancia es el valor [tex]t[/tex] del vértice de la parábola. Una forma rápida es mediante herramientas gráficas. Aquí encontramos que el granjero puede obtener ganancias máximas en el día 229.

Para aprender más sobre funciones cuadráticas, invitamos a ver esta pregunta verificada: https://brainly.com/question/8495268

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