If x^2+y^2=25,then log _5[max (3 x+4 y)] is | vedclass

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(A) Given $x^2+y^2=25$. Let $z = 3x + 4y$. By Cauchy-Schwarz inequality,for any real numbers $a, b, x, y$,we have $(ax + by)^2 \leq (a^2 + b^2)(x^2 +

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$2$

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(A) Given $x^2+y^2=25$. Let $z = 3x + 4y$. By Cauchy-Schwarz inequality,for any real numbers $a, b, x, y$,we have $(ax + by)^2 \leq (a^2 + b^2)(x^2 + y^2)$. Substituting $a=3, b=4$,we get $(3x + 4y)^2 \leq (3^2 + 4^2)(x^2 + y^2)$. $(3x + 4y)^2 \leq (9 + 16)(25) = 25 	imes 25 = 625$. Taking the square root,we get $-(25) \leq 3x + 4y \leq 25$. Thus,the maximum value of $3x + 4y$ is $25$. Therefore,$\log _5[\max (3x + 4y)] = \log _5(25) = \log _5(5^2) = 2 \log _5(5) = 2(1) = 2$.

If $x^2+y^2=25$,then $\log _5[\max (3 x+4 y)]$ is

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