How do you optimize $f(x,y)=2x^2+3y^2-4x-5$ subject to $x^2+y^2=81$ ?

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How do you optimize $f(x,y)=2x^2+3y^2-4x-5$ subject to $x^2+y^2=81$ ?

1 Answer

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Answer 1 · 2016-03-04T11:23:27

Maximum is 242, at $(-2, +-sqrt77 )$
There is no minimum.

Explanation:

Using $x^2+y^2=81$ ,
f(x, y) = g(x) = $238-x^2-4 x-5$
$d/dxg(x)=-2x-4 = 0$ , when $x=-2$ .
$d^2/dx^2g(x)--2<0$ .
When $x=-2,y=+-sqrt 77$ .
So, at $(-2,+-sqrt77)$ , f(x,y) attains its maximum value.
There is no other extreme value.

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How do you optimize $f(x,y)=2x^2+3y^2-4x-5$ subject to $x^2+y^2=81$ ?

1 Answer

Explanation:

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How do you optimize f(x,y)=2x^2+3y^2-4x-5f(x,y)=2x^2+3y^2-4x-5 subject to x^2+y^2=81x^2+y^2=81?

1 Answer

Explanation:

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How do you optimize $f(x,y)=2x^2+3y^2-4x-5$ subject to $x^2+y^2=81$ ?