Question

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

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

Let $z=3x+4y$
$\Rightarrow z=3x+4\sqrt{25-x^2}\left[\because x^2+y^2=25\right]$
$\therefore z^{\prime }=3+\frac{4}{2\sqrt{25-x^2}}(-2x)$
For maxima,put $z^{\prime }=0$
$\Rightarrow 3=\frac{4x}{\sqrt{25-x^2}}$
$\Rightarrow 3\sqrt{25-x^2}=4x$
$\Rightarrow 9\left(25-x^2\right)=16x^2$
$\Rightarrow 9\times 25-9x^2=16x^2$
$\Rightarrow 9\times 25=(16+9)x^2$
$\Rightarrow x^2=9$
$\therefore x=3,y=4$
Now, $z={\log }_5[(25)]={\log }_5(5{)}^2$
$=2{\log }_{5}5=2$