MATH HELP PLEASE!!!!!

GANDALF

Active member
i got this homework due tomorrow and i am stumped. ++karma for whoever can help me out.
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you have to set up two equations and then solve for one variable in one equation and plug it into the other

only in this problem there is profit you have to maximize so you have to plug the one variable in to the cost equation

you then take the derivative and set it equal to 0

this will give you your max and mins and then you can find the second derivative to prove that it is a max

simple max/ min calc really
 
Hes only in Math 2, unless that is calc I don't think it's that complicated. Regardless, ya that is what i did. I also changed it so that you multiplied everything for the chairs by 3, so that the profits were the same for the two different aspects, 1 table or 3 chairs. So you would have like 9 board feet and 6 labor hours for chairs.

9x+7y=420

6x+8y=400

multiply bottom equation by 3/2, so that it is 9x for both, subtract bottom from top and you are left with:

-6y=-180, you can take out a negative. Divide by 5, you get 36 for y. Plug that y value into either equation, and you get 56/3 for x. However, x is equal to chairs, and I multiplied the chair value times 3 earlier. So you could really say it is like, 3x+7y=420 or 2x+8y=400. This gives you a chair value of 56. Thus you have 36 tables (y value) and 56 chairs (x value), which will give you exactly 420 board feet and 400 hours labor. I didn't check to see if these are the max using calc, but I'm pretty sure it is just from how these equations work out.
 
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