Question

Let S={(x,y):x2+y2−6x−8y+21≤0}
Then max{12x7−5y7,(x,y)ϵS}+min{12(x2+y2+1)+(x−y);(x,y)ϵS}
−min{√3y+|x−3||x−3|;(x,y)ϵS} is

Answer 1

$S:(x-3{)}^2+(y-4{)}^2\le 4$

$\text{max}.\left\{\frac{12x-5y}{7}\right\}=\text{max}\{\frac{13}{7}\times \left(\frac{12x-5y}{13}\right)\}=\frac{13}{7}\text{max}\left\{\frac{12x-5y}{13}\right\}$ denotes max-distance of $(x,y)$ from line $12x-5y=0$


So, maximum distance$=2+\left|\frac{36-20}{13}\right|$
$=2+\frac{16}{13}=\frac{42}{13}$

so $\frac{13}{7}\text{max}\left\{\frac{12x-5y}{13}\right\}=\frac{13}{7}\times \frac{42}{13}=6\dots \left(1\right)$

for min $\left\{\frac{1}{2}\right(x^2+y^2+2x-2y+1\left)\right\}$

$=\frac{1}{2}\text{min}\left\{(x+1{)}^2+(y-1{)}^2-1\right\}$

denotes one less than the square of minimum distance between $(x,y)\&(-1,1)$

$=\frac{1}{2}({(\sqrt{(3+1{)}^2+(4-1{)}^2}-2)}^2-1)$

$=\frac{1}{2}(9-1)=4\dots \left(2\right)$

for min $\{\frac{\sqrt{3}y}{|x-3|}+1\}$

$\left\{\frac{y}{|x-3|}\right\}$denotes min slope of line joining $(x,y)\&(3,0)$

$\frac{y-0}{|x-3|}\to$ denotes min. slope when it is tangent from $(3,0)$ to the given circle.

which is equal to $tan\theta$.

$\tan \theta =\frac{2\sqrt{3}}{2}=\sqrt{3}$

so min$\left\{\frac{y}{|x-3|}\right\}=\sqrt{3}$

so$\sqrt{3}\left(\sqrt{3}\right)+1=4\dots \left(3\right)$
From (1),(2) & (3)

$6+4-4=6$