Help with math? Please

Discussion in 'Science and Nature' started by FlowerInTheSun, Jan 6, 2011.

  1. I"m sorry if this is in the wrong sub-forum...I really need some help though.
    This problem has been killing me for over 2 hours now.

    Farmer Fay has a pig that presently wieghs 200 pounds. She could sell it now for a price of $1.40 a pound. The pig is gaining 10 pounds a week while the price per pound is dropping 2 cents a week. When should Fay sell the pig to get the maximum amount of money for it.

    What I've done that has dead ended me several times now is this:
    I've made a chart which provided me with the quadratic equation:
    R= -2x^2-26x+280
    I then found my vertex which was at (-6.5, 98)
    (The x represents the numebr of weeks)

    What I have gathered (however incorrectly so) from this data is that if Farmer Fay waits 6.5 she will reach maximum profits. However, this being said when I checked it on the chart...The profits made at 6.5 weeks were not $98 nor was the maximum profit made.

    Help me PLEASE?! I really need some!
  2. #2 Spacelee, Jan 6, 2011
    Last edited by a moderator: Jan 6, 2011
    OK, I came up with this equation...
    f(x) = (200 + 10x)(1.40 - 0.02x)

    I'm only in AP calc, so don't trust my answer, maybe use it to help you or something lol idk.

    anyways... It's an optimization problem, or at least that's how I am solving it.
    f'(x) = 10 - 0.4x

    f'(x) = 0 @ x = 25

    f'(x).....+......0...... - ......

    so the slope of f(x) is increasing until it reaches 25 weeks (meaning the profit is increasing) and then after 25 weeks the slope is decreasing, thus proving it is a maximum.

    So your answer would be 25 weeks.
  3. What math class are you in? Seeing as it's a maximizing problem, i'd guess you would come up with a model equation then take the derivative of said equation and solve it when its equal to zero.

    I think the guy above me might have it
  4. Well I don't know if you're supposed to come up with some equation to show your teacher/professor, but the easiest way is this.

    200 x 1.40 = 280

    1 week:
    210 x 1.38 = 289.80

    2 weeks:
    220 x 1.36 = 299.20

    3 weeks:
    230 x 1.34 = 308.20

    4 weeks:
    240 x 1.32 = 316.80

    Ect ect ect.

    If you need an equation, this might work? I am not sure if i set it up right, but I used W as the variable for the weeks.

    Profit = (200 + (10W)) * (1.4 - (0.02W))

    Anyway, I'll leave you with that and that your answer is somewhere in the mid 20's.

    Edit: Ninja-ed. I take too long to do math these days, lol.

  5. i worked it out before seeing this post.. and this is what i got.

    99% sure it is correct :smoke:
  6. Thankyou guys sooo very much! This helped so much, you have noo idea :)
  7. yeha i just crunched the numbers and week 25 it 0.9 dollars a pound and 450 pounds

    love the optimization problem guy...well deserved plus rep coming your way...
  8. spacelee (or some other soul), can you also write out, in words, the above function?

    as in something like "the highest price obtainable for a currently 200 lbs.pig that gains 5% of its weight weekly in a current market of $1.40 per lbs. decreasing by 2 cents every week..."

    how does one put that equation into words?....i am trying to write it out here on paper but its not happening....*lemme toke more, maybe thatll open my creative passageways idk*...:smoke:
  9. I'm not exactly sure what to say (better at actually doing the problems then explaining things), but I'll do my best.

    f(x) = (200 + 10x)(1.40 - 0.02x)

    At the initial time we start the problem (x = 0), the pig weighed 200 pounds and the lady could sell the pig for 280$ at 1.40$ a pound. Every week the pig increases it's weight by 10 pounds and decreases it's price per pound by 2 cents. This keeps happening every week, but the price reaches a maximum after letting the pig gain weight for 25 weeks.

    P.S. I really don't know how I came up with the equation, I geuss I just found that weeks was a common variable and it looked like it had 2 parts to the equation with the weight and price.
  10. You come up with the equation by looking at it like this: You need to find how much the pig is worth right? In order to do that, you need to multiply the weight of the pig by the price per pound. So, the lets call these two variables x and y. So the equation is f(x)=xy. Now we find what x is and what y is. X (weight of pig) is found by this equation: (200 + 10x). Y (price per pound) is found by (1.40 - .02x). So put it all together and you get the final equation of: f(x)=(200 + 10x)(1.40 - 0.02x)
  11. thanks guy(edit:guys), i appreciate it...i guess i just dont have that natural synapses that make the click necessary in this kind of math...

    anyone into geometry?

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