Question

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 1

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 $= 125 + 3n \;\;kg \cdots(i)$

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 $= \ 5n \times 100 =\ 500n \cdots(ii)$

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 $= 1\times n = n$ cents.

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

By using equations (i) and (iii),

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

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

As 1 $ = 100 cents. so

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

Now, from equations (ii) and (iv)

Net profit, $P = 5(125+3n) (125-n) - 500n\cdots(v)$

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

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

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