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

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In summary, the minimum value of $(u-v)^2$ is 0, which occurs when u=v. The minimum value of $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ is 0, which occurs when u=1 and v=3. To find the minimum value of $(u-v)^2$, set u=v and then plug in the value of u into the equation. To find the minimum value of $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$, set u=1 and v=3 and then plug in the values into the equation. When u and v are not equal to each other, the
  • #1
anemone
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MHB
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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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  • #2
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 $(3-1)^2+(3-1)^2=8$.
 

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

1. What is the minimum value of $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$?

The minimum value of $(u-v)^2$ is 0, which occurs when $u=v$. The minimum value of $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ is also 0, which occurs when $u=\sqrt{2}$ and $v=3$.

2. How do you find the minimum value of $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$?

To find the minimum value of $(u-v)^2$, you can use the method of completing the square. For $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$, you can use the method of differentiation to find the critical points and determine the minimum value.

3. Can the minimum value of $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ be negative?

No, the minimum value of both $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ cannot be negative. This is because the square of any real number is always non-negative.

4. Are there any real values of $u$ and $v$ that make $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ equal to the minimum value?

Yes, as mentioned in the first question, the minimum value of both $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$ occurs when $u=v$ and $u=\sqrt{2}$, $v=3$, respectively.

5. What is the significance of finding the minimum value of $(u-v)^2$ and $\left(\sqrt{2-u^2}-\dfrac{9}{v}\right)^2$?

Finding the minimum value of a mathematical expression allows us to determine the smallest possible value that the expression can take. This can be useful in various applications, such as optimization problems in engineering and economics.

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