Friday

March 27, 2015

March 27, 2015

Posted by **anonymous** on Friday, February 11, 2011 at 4:32pm.

- Math app -
**anonymous**, Friday, February 11, 2011 at 6:17pmcould anyone help me please?

- Math app -
**Ms. Sue**, Friday, February 11, 2011 at 6:33pmThere'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.

- Math app -
**Anonymous**, Friday, February 11, 2011 at 7:52pmthe answers got to be 13wks and $96. but i couldn't get it.

- Math app -
**Ms. Sue**, Friday, February 11, 2011 at 8:12pmIf 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.

- Math app -
**tchrwill**, Saturday, February 12, 2011 at 6:44pmHow 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

- Math app -
**tchrwill**, Saturday, February 12, 2011 at 6:58pmSorry - 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.

**Answer this Question**

**Related Questions**

statistics - A random sample of 20 hogs on farmer Brown's hog farm had an ...

math : ( - would you like to brush your teeth with hogs hair? Toothbrushes were ...

math riddles for 7th grade - what was the truck driver doing w/a load of hogs

math - A typical Latino coffee farmer has fixed costs of $10,000 per year. Under...

Math - A farmer buys two varieties of animal feed. Type A contains ounces of ...

Elementary Algebra - The manager of a farmer's market has 500lb of grain that ...

algebra - Mixture Problems Template – Price Problems Copy the problem from the ...

Economic DQ's - 3. You have been hired by the government as an economic ...

math - Ivan bought 4.6 pounds of beef at the market for total of $36.57. He ...

CIS - Design a flowchart using Visual Logic for the following: A program that ...