SOLUTION: An orange grower has 400 crates of fruit ready to ship and will have 20 more for each day he waits. The current price is $60 per crate and will drop an estimated $2 per day for ea
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Question 21915: An orange grower has 400 crates of fruit ready to ship and will have 20 more for each day he waits. The current price is $60 per crate and will drop an estimated $2 per day for each day he waits. In how many days should he ship to maximize his income and what is his maximum income?
I tried to set up a equation like this: (400+20d)(60-2d)=i, but since it drops an estimated $2 a day, I'm don't think this is the right formula. Can you help? THANKS! Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! the profit is proportional to total income. we don't know the growers other
expenses he has to pay out.
my equation for total income is:
I = income
c = crates
#d = days passed
I = (400 c x $60 / c) + (20 c /d x #d)($60/c - ($2/d)/c x #d)
I = 400 x 60 + (20d)(60 - 2d)
I = 24000 + 1200d -40d^2
I = 2400 + 120 d - 4d^2
I = 600 + 30d - d^2
for a max or min I want I' the dirivative to be zero
0 = 30 - 2d
2d = 30
d = 15
I'll test for 14, 15 and 16 days to see if 15 days really is a maximum
I = 24,000 + 8,960
I = 24,000 + 8,960
I = 24,000 + 9,000
15 is a maximum
if you have not had calculus, you can use trial and error to get same result
