Answer

Let

y = a^2+b^2
(1)
3a+4b=5 \implies b = \dfrac{5-3a}{4}
(2)

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

Steven Zheng posted 22 hours ago

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