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## #1 2006-01-03 22:32:52

krisper
Member
Registered: 2005-12-20
Posts: 19

### inequality

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.

Humankind's inherent sense of right and wrong cannot be biologically explained.

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## #2 2006-01-04 02:57:55

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

### Re: inequality

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.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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## #3 2006-01-04 03:07:32

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: inequality

(x-7.5)^2+(y-5)^2 ≤ 3.25

IPBLE:  Increasing Performance By Lowering Expectations.

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## #4 2006-01-04 03:13:30

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: inequality

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)]

Last edited by krassi_holmz (2006-01-04 09:31:36)

IPBLE:  Increasing Performance By Lowering Expectations.

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## #5 2006-01-04 06:19:13

krisper
Member
Registered: 2005-12-20
Posts: 19

### Re: inequality

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.

Humankind's inherent sense of right and wrong cannot be biologically explained.

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