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An orange growe has 400 crates of fruit ready for maret andwill have 20 more each day the grower waits.
the present price is $60 per crate and will drop an estimated $2 per day for each day waited.
In how many days should the grower ship the crop for maximum income?
Let x = no. of days to wait for max income
Income = no. of crates * price/crate
f(x) = (400 + 20x)*(60 - 2x)
f(x) = 24000 - 800x + 1200x - 40x^2
Arrange as a quadratic equation
f(x) = -40x^2 + 400x + 24000
Max income occurs at the axis of symmetry, x = -b/(2a)
In this equation; a=-40; b= 400
x = +5 days for max income