who knew cow grains would be so difficult

Szafranko

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Joined
Apr 18, 2010
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6
talking about dividing polynomials and remainder theorem and what not....
i've yet to understand which values go where when talking about a word problem. its not as easy as jim has 2 applys sally wants 1.

so im not asking for a solve but perhaps just help which numbers are what and .. uh? lol a solve would be helpful as well since i have 3 pages of these to do haha

a 132kg cow grains 3.5 kg/day. the price of beef cattle is 5.10/kg, but the price falls by $ 0.10/day.
when should the cow be sold to maximize revenue

so obviously we have some type of graph going on here... umm other then that maybe .10 is the divider? and some fusino of how much the cow grains with beef price to create the dividend?
 
Szafranko said:
talking about dividing polynomials and remainder theorem and what not....
i've yet to understand which values go where when talking about a word problem. its not as easy as jim has 2 applys sally wants 1.

so im not asking for a solve but perhaps just help which numbers are what and .. uh? lol a solve would be helpful as well since i have 3 pages of these to do haha

a 132kg cow grains 3.5 kg/day. the price of beef cattle is 5.10/kg, but the price falls by $ 0.10/day.
when should the cow be sold to maximize revenue

so obviously we have some type of graph going on here... umm other then that maybe .10 is the divider? and some fusino of how much the cow grains with beef price to create the dividend?

I'll do a similar but different problem for you

a 150 kg cow grains 2 kg/day. the price of beef cattle is 125/kg, but the price falls by $ 1/day.
when should the cow be sold to maximize revenue

revenue = price * weight

Let the maximum revenue be after 'd' days

The weight of the cow after 'd' days = (150 + 2*d) kg

Price of beef = (125 - d) $/kg

Revenue = R = price * weight = (150 + 2*d)(125 - d)

If you plot "R" (y) vs. "d" (x) - you'll get a parabola with vertex at (25, 20000)

So the animal should be sold after 25days for total $20,000.
 
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