document.write( "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?\r
\n" ); document.write( "\n" ); document.write( "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!
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Algebra.Com's Answer #10612 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
the profit is proportional to total income. we don't know the growers other
\n" ); document.write( "expenses he has to pay out.
\n" ); document.write( "my equation for total income is:
\n" ); document.write( "I = income
\n" ); document.write( "c = crates
\n" ); document.write( "#d = days passed
\n" ); document.write( "I = (400 c x $60 / c) + (20 c /d x #d)($60/c - ($2/d)/c x #d)
\n" ); document.write( "I = 400 x 60 + (20d)(60 - 2d)
\n" ); document.write( "I = 24000 + 1200d -40d^2
\n" ); document.write( "I = 2400 + 120 d - 4d^2
\n" ); document.write( "I = 600 + 30d - d^2
\n" ); document.write( "for a max or min I want I' the dirivative to be zero
\n" ); document.write( "0 = 30 - 2d
\n" ); document.write( "2d = 30
\n" ); document.write( "d = 15
\n" ); document.write( "I'll test for 14, 15 and 16 days to see if 15 days really is a maximum\r
\n" ); document.write( "\n" ); document.write( "\"I+=+400+%2A+60+%2B+%2820+%2A+14%29%2860+-+%282+%2A+14%29%29\"
\n" ); document.write( "I = 24,000 + 8,960
\n" ); document.write( "\"I+=+400+%2A+60+%2B+%2820+%2A+16%29%2860+-+%282+%2A+16%29%29\"
\n" ); document.write( "I = 24,000 + 8,960
\n" ); document.write( "\"I+=+400+%2A+60+%2B+%2820+%2A+15%29%2860+-+%282+%2A+15%29%29\"
\n" ); document.write( "I = 24,000 + 9,000
\n" ); document.write( "15 is a maximum
\n" ); document.write( "if you have not had calculus, you can use trial and error to get same result
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