Question

At the beginning of the first day (day 1) after grape harvesting is completed, a grape grower has 8000 kg of grapes in storage. On day n, for n = 1, 2, . . . , the grape grower sells 250n/(n + 1) kg of the grapes at the local market at the price of $2.50 per kg. He leaves the rest of the grapes in storage where each day they dry out a little so that their weight decreases by 3%. Let wn be the weight (in kg) of the stored grapes at the beginning of day n for n ≥ 1 (before he takes any to the market).

(a) Find the value of wn for n = 2.

(b) Find a recursive definition for wn. (You may find it helpful to draw a timeline.)

(c) Let rn be the total revenue (in dollars) earned from the stored grapes from the beginning of day 1 up to the beginning of day n for n ≥ 1. Find a recursive formula for rn.

Answer 1

Solution:

Let's denote given values:

$w1$ =800 kg - total grapes beginning of the 1st day (before sell);

$k1=(250*1)/(1+1); k2=(250*2)/(2+1); .....k(n-1)=(250*(n-1))/((n-1)+1);$

and $kn=(250*n)/(n+1);$ weight of sold grapes 1st, 2nd ... and nth days (in kg).

$\alpha=0.03$ - rate of weight decreases by drying.

a) If $wn$ - weight of the stored grapes at the beginning of day nth, for n=2:

$w2=w1-k1-(w1-k1)*\alpha=(w1-k1)(1-\alpha);$

So,

$w2=(800kg-(250/2)kg)*(1-0.03)=654.75kg.$

b) We found for $w2=(w1-k1)(1-\alpha);$

so, for $w3=(w2-k2)(1-\alpha);$

if we continue this sequence, we can find:

$wn=(w(n-1)-k(n-1))(1-\alpha).$

c) We know:

$k1=(250*1)/(1+1); k2=(250*2)/(2+1); .....k(n-1)=(250*(n-1))/((n-1)+1);$

and $kn=(250*n)/(n+1);$ weight of sold grapes 1st, 2nd ... and nth days (in kg).

If $rn$ revenue up to beginning of the day n:

rn=k1*2.5$+k2*2.5$+...+k(n-1)*2.5$;

rn=250*2.5$(1/2+2/3+3/4+...+(n-1)/n).

Answer:

a) $w2=654.75 kg.$

b) $wn=(w(n-1)-k(n-1))(1-\alpha).$

c) rn=(1/2+2/3+3/4+...+(n-1)/n)*625$.