Question

Find the least value of the expression 4x2 + 2y2 – 4xy – 4y + 20

Answer 1

Answer:

4x2-4xy+y2+y^2-4y+4+16

=(2x-y)^2+(y-2)^2+16

since (2x-y^2> or =0

therefore least value is 0+0+16=16

Answer 2

Answer:

The least value of the expression 4x2 + 2y2 – 4xy – 4y + 20 is 16.

Step-by-step explanation:

4$x^{2}$ + 2$y^{2}$ - 4$x$$y$ - 4$y$ + 20

=$(2x)^{2}$ - 2*$2x$*$y$ + $y^{2}$ + $y^{2}$ - 2*$y$*2 + $2^{2}$ + 16

=$(2x-y)^{2} + (y-2)^{2} +16$

Here , the least value of any square is 0 .

So , the least value of  $(2x-y)^{2} and (y-2)^{2}$ are 0 .

Therefore , the least value of the expression 4x2 + 2y2 – 4xy – 4y + 20 is 16.

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