1. ## Farmer optimization question

Okay, I may be bad at optimization problems, but this question I can't even read it without getting a headache, and I have a test!!

If a farmer digs up his potatoes on July 1st, the crop is 120 m^3, which sell as new potatoes at $3.00/ m^3. If the crop is allowed to mature further, it increases at 15 m^3/ week. However, the price drops 20 cents/m^3/ week. When should the farmer dig the crop for the maximum cash return?? So am I supposed to find the profit or something or the day hes supposed to dig it up aghhhh 2. Originally Posted by Mathforum Okay, I may be bad at optimization problems, but this question I can't even read it without getting a headache, and I have a test!! If a farmer digs up his potatoes on July 1st, the crop is 120 m^3, which sell as new potatoes at$3.00/ m^3. If the crop is allowed to mature further, it increases at 15 m^3/ week. However, the price drops 20 cents/m^3/ week. When should the farmer dig the crop for the maximum cash return??

So am I supposed to find the profit or something or the day hes supposed to dig it up aghhhh
cash = (potato volume)(price/volume)

let t = number of weeks after July 1

set up the equation as a function of t ... let's see what you can do.

3. Okay, so to start you'd have the original amount, 120 x 3. Plus 15t minus 20 cents for each week.

4. So if the crop increases at 15m^3/ week...then thats \$3.00 more for each m^3...minus 20 cents for each m^3

5. C(t) = (360 + 15t) (3 - 20t) ?? am i on the right track

6. C(t) = (360 + 15t)(3- .20t)
= 1080 - 72t +45t - 3t^2
= -3t^2 - 27t- 1080
C'(t) = -6t - 27
Set C'(t) = 0
therforee... -6t -27 = 0

t = - 9/ 2 :S
That seems so wrong

7. Waiiit...I did the volume wrong hmm... C(t) = (120 + 15t) (3- .20t)

C(t) = -3t^2 + 21t +360
C'(t) = -6t +21
t = 3.5 weeks That seems better!!