A) 0
B) 1
C) No least value
D) none of these
Correct Answer: B
Solution :
(b)let \[f\left( x,y,z \right)={{x}^{2}}+4{{y}^{2}}+3{{z}^{2}}-2x-12y-6z+14\] \[={{\left( x-1 \right)}^{2}}+{{\left( 2y-3 \right)}^{2}}+3{{\left( z-1 \right)}^{2}}+1\] |
For least value of \[f\left( x,y,z \right)\] |
\[x-1=0;2y-3=0\,\operatorname{and}\,z-1=0\] |
\[\therefore x=1;y=\frac{3}{2};z=1\] |
Hence, least value of \[f\left( x,y,z \right)\operatorname{is}\,f\left( 1,\frac{3}{2},1 \right)=1\]. |
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