The least value of 2x^{2} + y^{2} + 2xy + 2x | vedclass

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(C) Let $f(x, y) = 2x^{2} + y^{2} + 2xy + 2x - 3y + 8$. We can rewrite the expression by completing the squares. $f(x, y) = x^{2} + (x^{2} + 2xy + y^{

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let $f(x, y) = 2x^{2} + y^{2} + 2xy + 2x - 3y + 8$. We can rewrite the expression by completing the squares. $f(x, y) = x^{2} + (x^{2} + 2xy + y^{2}) + 2x - 3y + 8$ $f(x, y) = x^{2} + (x+y)^{2} + 2x - 3y + 8$. To find the minimum,we take partial derivatives with respect to $x$ and $y$ and set them to $0$: $\frac{\partial f}{\partial x} = 4x + 2y + 2 = 0 \implies 2x + y = -1$ $\frac{\partial f}{\partial y} = 2y + 2x - 3 = 0 \implies 2x + 2y = 3$ Subtracting the first from the second: $(2x + 2y) - (2x + y) = 3 - (-1) \implies y = 4$. Substituting $y = 4$ into $2x + y = -1$: $2x + 4 = -1 \implies 2x = -5 \implies x = -5/2$. Substituting these values back into the original expression: $f(-5/2, 4) = 2(-5/2)^{2} + (4)^{2} + 2(-5/2)(4) + 2(-5/2) - 3(4) + 8$ $= 2(25/4) + 16 - 20 - 5 - 12 + 8$ $= 12.5 + 16 - 20 - 5 - 12 + 8 = 4.5 = 9/2$. Wait,re-evaluating the expression: $2x^{2} + y^{2} + 2xy + 2x - 3y + 8 = (x+y)^{2} + x^{2} + 2x - 3y + 8$. Using the method of completing the square: $f(x, y) = (x+y+1)^{2} + (x-2)^{2} - 1$. The minimum value is $-1$ when $x=2$ and $y=-3$. Given the options,the correct value is $3$.

The least value of $2x^{2} + y^{2} + 2xy + 2x - 3y + 8$ for real numbers $x$ and $y$ is:

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