Quadratic Functions ALGEBRA? - value calculator to sell beanie babies
Quadratic algebra?
In production, the manufacture of an electronics company projects the following costs for a new series of computers. Marketing believes that the best retail price is $ host 92nd
1. Assuming that this relationship is quadratic, you can find the quadratic function C (x) = ax2 + bx + c to adapt
these data.
2. If each simulator sold $ 92, and X is the number of computers sold, enter the income
Function?
3. Profit is revenue minus operating costs. Enter the profit equation.
4. How many computers are sold, this line) Benefits Calculator (which values of x positive benefits?
5. How many computers are sold to maximize profit? In other words, what value of x will benefit the generalis?
6. How can you verify whether the work? (Tip: Prepare this on a table, to check his work.)
Number of computers / calculators cost x
0 / $ 92,000
10 / $ 84,200
100 / $ 32,000
Can anyone me how these problems? I have some problems with this kind of question. This is a question of me and my final exam in a week! Thank you!
Wednesday, February 17, 2010
Value Calculator To Sell Beanie Babies Quadratic Functions ALGEBRA?
1 comment:
Sorry, but how can it cheaper at 100 computer 10 to produce?
It makes no sense. And these figures should not the cost of the computer as well as the cost for 0 would be infinite. I assume that there instead of 0, is a 1 in the above table and the cost is the cost for the computer book.
F For all three pairs here (1.92000), (10.84200), (100.32000) you can find a quadratic function f, so that three pairs are given by (x1, (x1)), (x2, f (x2)) and (x3, f (x3)).
Consider the function g (x) = (x-x1) * (x-x2). She disappears in X1 and X2, and only there. So if you have a function h that the correct values for x1 and x2, but not the correct value for X3, a suitable multiple of G to H might add, to obtain the desired function f. As we have found,Hours from now.
Suppose you have two lines through the points (x 1, 1), (x2, y2), or resp. (x1, y1) (x2, 1). If you could multiply these two functions h.
Call these affine functions L and K. We
l (x) = 1 + ((y2-1) / (x2-x1)) * (x-x1) and
K (x) = 1 + ((y1-1) / (X1-X2)) * (x-x2).
In this example, the
l (x) = 1 + ((84200-1) / 9) * (x-1) = 1 + (84199 / 9) * (x-1) = -84190 / 9 + (84199 / 9) * x
and
K (x) = 1 + (92000-1) / -9) * (x-10) = 1 - (91999 / 9) * (x-10) = 919999 / 9 - (91999 / 9) * x
So
h (x) = L (x) * k (x) = - (84199 * 91999 / 81) x ^ 2 + (84190 * 91999 * 84199 + 919999) / 81) x - 84190 * 919999 / 81
Now you need to h (100 to be payable) and see what it is.If there is 32000, and f = h, otherwise we have H + F = G * (32000 - H (100)) / g (100)
I hope that helps.
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