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Least value of expression x^2+4y^2+3z^2-4x-20y-12z+45 is.?

1 Answer

7 votes
Hello,

4 is the least value.

Indeed:
x²-4x+4=(x-2)²
4y²-20y+25=(2y-5)²
3(z²-4z+4)=3(z-2)²

x²+4y²+3z²-4x-20y-12z+45= (x-2)²+(2y-5)²+3(z-2)²+45-4-25-12
=(x-2)²+(2y-5)²+3(z-2)²+4

Minimum is 4 when x=2,y=5/2 and z=2 for square is always positive.

answered
User StillLearning
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