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Math Help - Quadratic Equations

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    Quadratic Equations

    Find the smallest possible value of the quantity x^2 + y^2 under the restriction that 2x + 3y = 6

    How would I go about solving this? Please help.
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  2. #2
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    Quote Originally Posted by lightningstab714 View Post
    Find the smallest possible value of the quantity x^2 + y^2 under the restriction that 2x + 3y = 6

    How would I go about solving this? Please help.
    2x + 3y = 6 \Rightarrow x = \frac{6 - 3y}{2} .... (1)

    Substitute (1) into L = x^2 + y^2 and find the value of y that minimises L. Substitute this value into L to get the minimum value of L.
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    Senior Member pankaj's Avatar
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    Here are two non calculus approaches to your problem:

    Geometrically, x^2+y^2 denotes square of distance of the point (x,y) from (0,0) therefore,you have been asked to find the shortest distance between (0,0) and the line 2x+3y=6 in 2D space which is the perpendicular distance of 2x +3y=6 from (0,0) and which can be found from the formula
    p=\frac{|ax_{1}+by_{1}+c|}{\sqrt{a^2+b^2}}
    =\frac{|2(0)+3(0)+6|}{\sqrt{2^2+3^2}}
    =\frac{6}{\sqrt{13}}.Therefore answer should be <br />
\frac{36}{13}

    Another method:

    Put x=r\cos\theta;y=r\sin \theta where r=\sqrt{x^2+y^2}

    Plugging in 2x+3y=6,we get

    r=\frac{6}{2\cos \theta +3\sin \theta}

    Now r will be minimum when denominator will be maximum.

    Also ,-\sqrt{a^2+b^2} \leq a\cos\theta+b\sin\theta \leq\sqrt{a^2+b^2}

    Therefore, r_{min}=\frac{6}{\sqrt{2^2+3^2}}=\frac{6}{\sqrt{13  }}
    and hence x^2+y^2=\frac{36}{13}
    Last edited by pankaj; September 19th 2008 at 06:59 PM.
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