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)]]]>

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.

]]>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.

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