Question

If $x^{2}+y^{2}=25$, then $\log _{5}[\max (3 x+4 y)]$ is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

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