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anemone
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Find the minimum value of $(x-y)^2+\left( \sqrt{2-x^2}-\dfrac{9}{y} \right)^2$ for $0<x<\sqrt{2}$ and $y>0$.
anemone said:Find the minimum value of $(x-y)^2+\left( \sqrt{2-x^2}-\dfrac{9}{y} \right)^2$ for $0<x<\sqrt{2}$ and $y>0$.
The purpose is to determine the lowest possible value that the expression can have, which can be useful in optimization problems or in finding the equilibrium point of a system.
To find the minimum value, we can use techniques such as differentiation or completing the square and setting the derivative or completed expression equal to 0.
No, the minimum value cannot be negative because it is the square of a real number, and squares are always non-negative.
Yes, there are restrictions on the values of x and y. For the expression to be defined, x must be between -√2 and √2, and y must be non-zero.
Yes, it is possible for the minimum value to be achieved at multiple points, especially if the expression contains absolute values or has multiple local minimums.