SOLUTION: an orange grower has 400 crates or fruit ready and will have 20 more for each day the grower waits to sell. the present price is $60 per crate and will drop $2 per crate per day. i
Question 181439: an orange grower has 400 crates or fruit ready and will have 20 more for each day the grower waits to sell. the present price is $60 per crate and will drop $2 per crate per day. in how many days should the grower ship the crop to maximize income? Found 2 solutions by nerdybill, stanbon:Answer by nerdybill(7384) (Show Source):
You can put this solution on YOUR website! .
Let x = number of days until shipped
then
f(x) = (60-2n)(400 + 20n)
where f(x) is the income
.
f(x) = (60-2n)(400 + 20n)
f(x) = 24000+1200n-800n-40n^2
f(x) = -40n^2+400n+24000
.
Since the coefficient for the n^2 term is -40 (negative), we know that it is a parabola the opens downward. So finding the axis of symmetry will give us the max.
.
axis of symmetry = -b/2a = -400/2(-40) = -400/-80 = 5
.
Answer: 5 days
You can put this solution on YOUR website! an orange grower has 400 crates of fruit ready and will have 20 more for each day the grower waits to sell. the present price is $60 per crate and will drop $2 per crate per day. in how many days should the grower ship the crop to maximize income?
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Revenu = price * quantity
R(x) = (60 - 2x)(400 + 20x)
R(x) = 2(30-x)*20(20+x)
R(x) = -40(x-30)(x+20)
R(x) = -40(x^2 - 10x - 600)
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Max of min occurs when x = -b/2a = 10/2 = 5
Maximum revenue will occur on day 5
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Cheers,
Stan H.
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