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Math Help - minimum value

  1. #1
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    Smile minimum value

    hi
    let x and y two reels such that, (x+5)^2+(y-12)^2=14^2,find the minimum value of x^2+y^2.
    thanks.
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  2. #2
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    Quote Originally Posted by Raoh View Post
    hi
    let x and y two reels such that, (x+5)^2+(y-12)^2=14^2,find the minimum value of x^2+y^2.
    thanks.
    1. The first equation describes a circle around M(-5, 12) with r = 14.
    The 2nd term belongs to a circle around the origin with a variable radius.

    2. The minimum of x + y = r occurs when the 2nd circle is tangent to the first circle. In this case both midpoints and the tangent point are placed on a straight line through the origin (= center of the 2nd circle)

    3. Determine the tangent point by solving for (x, y):

    \left|\begin{array}{rcl}(x+5)^2+(y-12)^2&=&14^2 \\ y &=& -\frac{12}5 x \end{array} \right.

    4. I've got the minimum distance d_{min} = 1
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    thank you.
    i have another approach i hope you would help me finish it.
    let x+5=14\cos(\phi) and y-12=14\sin(\phi) such that \phi\in [0,2\pi )
    i ended up with,
    x^2+y^2=\left (14\cos(\phi )-5  \right )^2+(12+14\sin(\phi))^2 =196 \cos^2(\phi)-140 \cos(\phi)+25+196 \sin^2(\phi)+336 \sin(\phi)+144= 196-140 \cos(\phi)+336 \sin(\phi)+169=365-140 \cos(\phi)+336 \sin(\phi)
    any ideas how to proceed ?
    thanks.
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  4. #4
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    Here's yet another way to do that: x^2+ y^2 is the square of the distance from (0, 0) to (x, y) and will be minimized when the distance is- and the shortest distance is always on a straight line. Draw a Straight line from (0, 0) to (-5, 12), center of the circle. The point on the circle closest to (0, 0) will be the point where that line crosses the circle. Find the equation of that line (since it passes through (0, 0) it can be written as y= mx) and solve that and the equation of the circle. That will result in a quadratic which has two solutions- one solution will be point that minimizes x^2+ y^2, the other the point where it is maximized.
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  5. #5
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    Quote Originally Posted by Raoh View Post
    thank you.
    i have another approach i hope you would help me finish it.
    let x+5=14\cos(\phi) and y-12=14\sin(\phi) such that \phi\in [0,2\pi )
    i ended up with,
    x^2+y^2=\left (14\cos(\phi )-5  \right )^2+(12+14\sin(\phi))^2 =196 \cos^2(\phi)-140 \cos(\phi)+25+196 \sin^2(\phi)+336 \sin(\phi)+144= 196-140 \cos(\phi)+336 \sin(\phi)+169=365-140 \cos(\phi)+336 \sin(\phi)
    any ideas how to proceed ?
    thanks.
    Far from the easiest way! But now that you have it, look for the value of \phi that minimizes the function by differentiating it with respect to \phi and setting that derivative equal to 0. It will eventually reduce to tan(\phi)= some constant, again telling you that the extremal points lie on a straight line from the origin.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Far from the easiest way! But now that you have it, look for the value of \phi that minimizes the function by differentiating it with respect to \phi and setting that derivative equal to 0. It will eventually reduce to tan(\phi)= some constant, again telling you that the extremal points lie on a straight line from the origin.
    thank you "Hallsofivy"
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    Quote Originally Posted by HallsofIvy View Post
    Here's yet another way to do that: x^2+ y^2 is the square of the distance from (0, 0) to (x, y) and will be minimized when the distance is ..?.. and the shortest distance is always on a straight line. Draw a Straight line from (0, 0) to (-5, 12), center of the circle. The point on the circle closest to (0, 0) will be the point where that line crosses the circle. Find the equation of that line (since it passes through (0, 0) it can be written as y= mx) and solve that and the equation of the circle. That will result in a quadratic which has two solutions- one solution will be point that minimizes x^2+ y^2, the other the point where it is maximized.
    thanks but i think there's something missing.
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  8. #8
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    Quote Originally Posted by Raoh View Post
    thank you.
    i have another approach i hope you would help me finish it.
    let x+5=14\cos(\phi) and y-12=14\sin(\phi) such that \phi\in [0,2\pi )
    i ended up with,
    x^2+y^2=\left (14\cos(\phi )-5  \right )^2+(12+14\sin(\phi))^2 =196 \cos^2(\phi)-140 \cos(\phi)+25+196 \sin^2(\phi)+336 \sin(\phi)+144= 196-140 \cos(\phi)+336 \sin(\phi)+169=365-140 \cos(\phi)+336 \sin(\phi)
    any ideas how to proceed ?
    thanks.
    1. All your calcualtions are correct!

    2. You have now a function in \phi of which you want to get a minimum: Let s(\phi) denote the function:

    s(\phi)=365-140 \cos(\phi)+336 \sin(\phi)

    Differentiate wrt \phi and solve the equation s'(\phi) = 0 for \phi

    3.
    336 \cos(\phi) + 140 \sin(\phi) = 0

    4. Divide the equation by \cos(\phi) and keep in mind that \frac{\sin(\phi)}{\cos(\phi)} = \tan(\phi)

    5. Plug in the result of \phi into s(\phi).
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    Here's yet another way to do that: x^2+ y^2 is the square of the distance from (0, 0) to (x, y) and will be minimized when the distance is
    No, that's perfectly correct English: "The square of the distance will be minimized when the distance is". That's equivalent to "The square of the distance will be minimized when the distance is minimized".

    - and the shortest distance is always on a straight line. Draw a Straight line from (0, 0) to (-5, 12), center of the circle. The point on the circle closest to (0, 0) will be the point where that line crosses the circle. Find the equation of that line (since it passes through (0, 0) it can be written as y= mx) and solve that and the equation of the circle. That will result in a quadratic which has two solutions- one solution will be point that minimizes x^2+ y^2, the other the point where it is maximized.
    Quote Originally Posted by Raoh View Post
    thanks but i think there's something missing.
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