farmer Linton wants to find the best time to take her hogs to market. The current price is 88cents per pound, and her hogs weigh an average of 90 pounds. the hogs gain 5 pounds per week, and the market price for hogs is falling each week by 2 cents per pound. How many weeks should Ms. Linton wait before taking her hogs to market in order to receive as much money as possible? at the time, how much money (per hog) will she get?
6 answers
could anyone help me please?
There's probably a formula for figuring this out -- but I don't know it. A math tutor may be on later who can help.
I've been working with trial and error -- and I'm up to 9 weeks when she'd get $94.50 for a 135-pound hog. You can keep working with trial and error.
Your problem doesn't state what her costs are for feeding these hogs until she can reach the maximum price.
I've been working with trial and error -- and I'm up to 9 weeks when she'd get $94.50 for a 135-pound hog. You can keep working with trial and error.
Your problem doesn't state what her costs are for feeding these hogs until she can reach the maximum price.
the answers got to be 13wks and $96. but i couldn't get it.
If you count the original weight of 90 pounds and $0.88 as the first week, then the 13th week is 150 pounds at $0.64, then the price is $96.
By my calculations, week 14 would see a 155-pound hog at $0.62 = $96.10
The question is terribly flawed because it doesn't take into consideration the costs of feeding the hogs.
By my calculations, week 14 would see a 155-pound hog at $0.62 = $96.10
The question is terribly flawed because it doesn't take into consideration the costs of feeding the hogs.
How many weeks should Ms. Linton wait before taking her hogs to market in order to receive as much money as possible?
As the weight increases, the price decreases, the maximum income to be derived when the product of the weight and price is maximum.
We know that the weight W = 90 + 5w and the price P = 88 - 2w, w being the number of weeks to maximum income.
W = 90 + 5w
P = 88 - 2w
W(P) = (90 + 5w)(88 - 2w) yielding
w^2 - 26w - 792
Eq
As the weight increases, the price decreases, the maximum income to be derived when the product of the weight and price is maximum.
We know that the weight W = 90 + 5w and the price P = 88 - 2w, w being the number of weeks to maximum income.
W = 90 + 5w
P = 88 - 2w
W(P) = (90 + 5w)(88 - 2w) yielding
w^2 - 26w - 792
Eq
Sorry - I hit the send key by mistake.
How many weeks should Ms. Linton wait before taking her hogs to market in order to receive as much money as possible?
As the weight increases, the price decreases, the maximum income to be derived when the product of the weight and price is maximum.
We know that the weight W = 90 + 5w and the price P = 88 - 2w, w being the number of weeks to maximum income.
W = 90 + 5w
P = 88 - 2w
W(P) = (90 + 5w)(88 - 2w) yielding
w^2 - 26w - 792
Equating to zero and taking the first derivitive yields 2w - 26 = 0 making w = 13 weeks.
At the start of week #1, the weight = 90 lbs. and the price = 88 cents per pound. At the end of week 13, the weight is 90 + 5(13) = 155 lbs. and the price is 88 - 2(13) = 62 cents per pound.
Thus, during the 14th week, the hogs are sold on the basis of a weight of 155 lbs. and a selling price of 155(.62) = $96.10 each.
How many weeks should Ms. Linton wait before taking her hogs to market in order to receive as much money as possible?
As the weight increases, the price decreases, the maximum income to be derived when the product of the weight and price is maximum.
We know that the weight W = 90 + 5w and the price P = 88 - 2w, w being the number of weeks to maximum income.
W = 90 + 5w
P = 88 - 2w
W(P) = (90 + 5w)(88 - 2w) yielding
w^2 - 26w - 792
Equating to zero and taking the first derivitive yields 2w - 26 = 0 making w = 13 weeks.
At the start of week #1, the weight = 90 lbs. and the price = 88 cents per pound. At the end of week 13, the weight is 90 + 5(13) = 155 lbs. and the price is 88 - 2(13) = 62 cents per pound.
Thus, during the 14th week, the hogs are sold on the basis of a weight of 155 lbs. and a selling price of 155(.62) = $96.10 each.