Minimum Value of $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$

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SUMMARY

The minimum value of the expression $(u-v)^2+\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ for the constraints $00$ can be determined through calculus and optimization techniques. The problem involves analyzing the behavior of the function within the specified domain and applying methods such as the Lagrange multipliers or substitution to simplify the expression. The solution reveals that the minimum occurs at specific values of $u$ and $v$, which can be calculated explicitly.

PREREQUISITES
  • Understanding of calculus, particularly optimization techniques.
  • Familiarity with algebraic manipulation and inequalities.
  • Knowledge of the properties of square roots and their behavior in real numbers.
  • Experience with Lagrange multipliers for constrained optimization problems.
NEXT STEPS
  • Study optimization techniques using Lagrange multipliers in depth.
  • Explore the properties of square root functions and their derivatives.
  • Learn about constrained optimization problems in calculus.
  • Practice solving similar problems involving minimization of quadratic expressions.
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Mathematicians, students studying calculus, and anyone interested in optimization problems in real analysis.

anemone
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Here is this week's POTW:

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Find the minimum value of $(u-v)^2+\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ for $0<u<\sqrt{2}$ and $v>0$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered last week's POTW.(Sadface)

Below is a suggested solution:
The given function is the square of the distance between a point of the quarter of circle $x^2+y^2=2$ in the open first quadrant and a point of the half hyperbola $xy=9$ in that quadrant. The tangents to the curves at (1, 1) and (3, 3) separate the curves, and both are perpendicular to $x=y$, so those points are at the minimum distance, hence the answer is $(3-1)^2+(3-1)^2=8$.
 

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