Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is A $2$

Lets take $x=5\cos \theta ,y=5\sin \theta$
then $Max(3x+4y)=Max\left[5\right(3\cos \theta +4\sin \theta \left)\right]$
let $\sin x=\frac{3}{5}$ and $\cos x=\frac{4}{5}$
$=Max\left[25\sin (\theta +x)\right]=25$
So, ${\log }_5\left[Max\right(3x+4y\left)\right]={\log }_{5}25=2$
Or
Let $f(x,y)=3x+4y$ be a function. So we have to find maxima of $f(x,y)$ if $x^2+y^2=25$.
$f=3x+4\sqrt{25-x^2}$
For maxima $f^{\prime }\left(x\right)$ must be $0$.
$\Rightarrow 3-\frac{4x}{\sqrt{25-x^2}}=0$
$\Rightarrow x^2=9$
$\Rightarrow x=\pm 3$
and so $y=\pm 4$
Here we can see $f(x,y)$ is max at $(3,4)$ and min at $(-3,-4)$
So, ${\log }_5\left[Max\right(3x+4y\left)\right]={\log }_{5}25=2$