Question

A strawberry farmer will receive 37 per bushel of strawberries during the first week of harvesting each week after that the value will drop 0.8 per bushel the farmer estimates that there are approxiamately 129 of strawberries int he fieldsand that the crop is increasing at a rate of four bushels per week when should the farmer harvest the strawberries to maximize their value? (Assume that "during the first week of harvesting" here means week 1. How many bushels of strawberries will yield the maximum value? (Round your answer to the nearest integer.)

bushels
2. What is the maximum value of the strawberries? (Round your answer to two decimal places.)
$

Answer 1

a.)

Number of bushes = b = 129 + 4(7) = 129 + 28 = 157 bushes

b.)

Revenue = bx = (157)(31.4) = $ 4929.8

Explanation:

Let number of bushes = b

The price of the bushes = x

The time to harvest = t

The revenue = y

Now given,

Price = x = 37 - (0.8)t

and

Number of bushes = b = 129 + 4t

Also,

The revenue = (Number of bushes)(Price of the bushes)

= bx

= (129 + 4t)[37 - (0.8)t]

= 129×37 - 129× (0.8)t + 4×37t - 4×0.8t²

= 4773 - 103.2t + 148t - 3.2t²

= 4773 + 44.8t - 3.2t²

⇒ y = 4773 + 44.8t - 3.2t²

Now,

`(dy)/(dt)` = 0 + 44.8 - 6.4t

Now,

`(dy)/(dt)` = 0

⇒ 44.8 - 6.4t = 0

⇒ - 6.4t = -44.8

⇒ 6.4t = 44.8

⇒t =
`(44.8)/(6.4) = 7`

Now,

`(d^(2)y )/(dt^(2) )` = -6.4

As
`(d^(2)y )/(dt^(2) )` < 0 ,

It means at time t = 7 , the revenue is maximum

a.)

Number of bushes = b = 129 + 4(7) = 129 + 28 = 157 bushes

Price = x = 37 - (0.8)(7) = 37 - 5.6 = 31.4

b.)

Revenue = bx = (157)(31.4) = 4929.8