262put z=x+iy
N =numerator = { 2x2+(x-3)2 } + {2y2 +(y-6)2 }
for minimum value , \frac{\partial N}{\partial x} = \frac{\partial N}{\partial y} =0
from the above relation we get , x=1 y=2
N = 2+4+8+16 =30
N/10 =3 (Ans)
1708
\hspace{-16}$Here We have to find Min. of $\mathbf{\left(\mid z \mid^2+\mid z-3 \mid^2+\mid z-6i \mid^2\right)}$\\\\ Now Put $\mathbf{z=x+iy},$ We Get $\mathbf{\{(x^2+y^2)+(x-3)^2+y^2+x^2+(y-6)^2\}}$\\\\ Using Geometrically Let $\mathbf{A(0,0)\;,B(0,6)\;,C(3,0)}$\\\\ Which Shows a $\mathbf{\triangle ABC}$ and $\mathbf{G(x,y)}$ be any point\\\\ Now $\mathbf{GA^2+GB^2+GC^2}$ is Min., Whin $\mathbf{G}$ is a Centroid of $\mathbf{\triangle ABC}$\\\\ So $\mathbf{GA^2+GB^2+GC^2=30}$\\\\ So $\mathbf{Min.\frac{1}{10}\{GA^2+GB^2+GC^2\}=\frac{30}{10}=3}$
1MY solution is exactly as man111 the generalised equation is this
PA^2+PB^2+PC^2=GA^2+GB^2+Gc^2+3PG^2 (P.s this equation is too helpful it can be used in a lot of complex number solution)
1
varun.tinkle
·2012-03-13 23:20:01
Alternatively write
this
as 3x^2+3y^2-6x_12y+45/10
=3/10(x^2+y^2-2x-4y+15)
now write it
3/10((x-1)^2+(y-2)^2 ) obviously the minimum is reache dwhen x=1 y=2
1
samagr@
·2012-03-15 14:23:27
@aditya , for minima, your condition for PARTIAL DERIVATIVE is incomplete.you need to check the signs for higher orders also : ∂2N/∂x2, ∂2N/∂y2, ∂2N/∂x∂y
.
btw nice solution varun
262
Aditya Bhutra
·2012-03-15 20:22:56
i left that because it was obvious after that.
