If $$3a+4b=5$$, then what is the minimum value of $$ a^2+b^2$$

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Uniteasy Admin · Accepted Answer

Let 
$$y = a^2+b^2$$n
$$3a+4b=5 \implies b = \dfrac{5-3a}{4} $$n
Then
$$y = a^2+\Big(  \dfrac{5-3a}{4}\Big) ^2$$
$$=\dfrac{1}{16}(25a^2-30a+25) $$
$$=\dfrac{25}{16}(a^2-\dfrac{6}{5}a+1 ) $$
$$=\dfrac{25}{16}\Big[ \Big( a-\dfrac{3}{5} \Big) ^2+\dfrac{16}{25} \Big] $$
Therefore
$$y_{min} = 1,$$ when $$a = \dfrac{3}{5}$$
The minimum value of $$ a^2+b^2$$ is equal to 1

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