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krisper
2006-01-05 05:19:13

This is exactly where I am, but I just cant figure out how to use that. I guess I have to express x with y but just dont know how.

krassi_holmz
2006-01-05 02:13:30

This may help:
A^2+B^2 ≤C
0 ≤ A^2 ≤ C-B^2, SO
C-B^2 >= 0
-B^2 >= -C
B^2 <= C
B ∈ [-√C, +√C] and A  ∈ [-√(C-B^2), +√(C-B^2)]

krassi_holmz
2006-01-05 02:07:32

(x-7.5)^2+(y-5)^2 ≤ 3.25

Ricky
2006-01-05 01:57:55

First, take the equation:

x(x-15) + y(y-10) + 78 <= 0

and complete the square.  You should end up with an equation:

(x-a)^2 + (y-b)^2 <= r^2

where a, b, and r are constants.

Now how much calculus do you know?  Since you are finding min and max's, I have to assume you know some, but this seems like a Multivariable Calculus question.  Is this what you're taking?

If so, all you need to do is derivative with respect to x and y, find the zeros and the end points on the region.  Then find which one of these is the highest and which is the lowest.

krisper
2006-01-04 21:32:52

For y and x we have this : x(x-15) + y(y-10) + 78 <= 0; Find the max and the min value of the expression : 3x + 2y.

I dont know if this is the right method for this problem but here is where I am:

I. x^2 - 15x + y^2 - 10y + 78 <= 0
5(3x + 2y) => x^2 + y^2 + 78
And here I dont know what to do so I try another method:

II.  x^2 - 15x + y^2 - 10y + 78 <= 78
x^2 - 15x + 225/4 <= 10y - y^2 - 25 + 13/4
(x-7,5)(x+7,5) <= -(y-5)(y-5) + 13/4
(x-7,5)(x+7,5) <= (√13/2 - y + 5)(√13/2 + y - 5)
(x-7,5)(x+7,5) <= 1/4(√13 - 2y + 10)(√13 + 2y - 10)

But here I am stuck again. I find the roots of these 2 equasions for x and y, draw their parabolas and compare them but the only thing I find is an interval of solutions not min and max values of the 3x + 2y. Please help Thanks.