How do you optimize $f (x, y) = x^{2} - 3 y^{2}$ $f(x,y)=x^2-3y^2$ subject to $x^{3} - y^{2} = 8$ $x^3-y^2=8$ ?

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How do you optimize $f (x, y) = x^{2} - 3 y^{2}$ $f(x,y)=x^2-3y^2$ subject to $x^{3} - y^{2} = 8$ $x^3-y^2=8$ ?

1 Answer

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Answer 1 · 2016-03-04T11:55:42

There is no finite extreme value.

Explanation:

Using $y^{2} = x^{3} - 8$ $y^2=x^3-8$ ,
f(x, y) = g(x) = $- 3 x^{3} + x^{2} + 24$ $-3x^3+x^2+24$ '
$\frac{d}{d x} g (x) = - 9 x^{2} + 2 x = 0 \to x = 0 or \frac{2}{9}$ $d/dxg(x)=-9x^2+2x=0 to x =0 or 2/9$
$y^{2} < 0$ $y^2 < 0$ ,
So, these points (x, y) are unreal.
For y to be real, $x \geq 2$ $x>=2$ ..

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How do you optimize $f (x, y) = x^{2} - 3 y^{2}$ $f(x,y)=x^2-3y^2$ subject to $x^{3} - y^{2} = 8$ $x^3-y^2=8$ ?

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How do you optimize f(x,y)=x2−3y2f(x,y)=x^2-3y^2 subject to x3−y2=8x^3-y^2=8?

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Explanation:

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How do you optimize $f (x, y) = x^{2} - 3 y^{2}$ $f(x,y)=x^2-3y^2$ subject to $x^{3} - y^{2} = 8$ $x^3-y^2=8$ ?