, 2016

Solution 1

We expand the given expression to obtain $$(x^2+6x+9) + 2(y^2 - 4y + 4) + 4(x^2 - 14x + 49) + (y^2 + 8y + 16)$$ We expand further to obtain $$x^2+6x+9 + 2y^2 - 8y + 8 + 4x^2 - 56x + 196 + y^2 + 8y + 16$$ We simplify to obtain $$5x^2 - 50x + 3y^2 + 229$$ We remove a common factor of 5 from the first two terms $$5(x^2 - 10x) + 3y^2 + 229$$ and then complete the square to obtain $$5(x^2 - 10x + 5^2 - 5^2) + 3y^2 + 229$$ This gives $$5(x-5)^2 - 125 + 3y^2 + 229$$ or $$5(x-5)^2 + 3y^2 + 104$$ Since $(x-5)^2 \geq 0$ for all real numbers $x$ and $3y^2 \geq 0$ for all real numbers $y$, then the minimum value of $5(x-5)^2 + 3y^2 + 104$ (and hence of the original expression) is $5(0)+3(0)+104$ or 104.

We note that this value is actually achieved when $x = 5$ (which gives $(x-5)^2 = 0$) and $y=0$ (which gives $3y^2 = 0$).

Solution 2

We expand the given expression to obtain $$(x^2+6x+9) + 2(y^2 - 4y + 4) + 4(x^2 - 14x + 49) + (y^2 + 8y + 16)$$ We expand further to obtain $$x^2+6x+9 + 2y^2 - 8y + 8 + 4x^2 - 56x + 196 + y^2 + 8y + 16$$ The terms involving $x$ are $$x^2+6x+9+ 4x^2 - 56x + 196 = 5x^2 - 50x + 205 = 5x^2-50x+125+80 = 5(x-5)^2 + 80$$ The terms involving $y$ are $$2y^2 - 8y + 8 + y^2 + 8y + 16 = 3y^2 + 24$$ Since $(x-5)^2 \geq 0$ for all real numbers $x$, then the minimum value of $5(x-5)^2 + 80$ is 80.

Since $y^2 \geq 0$ for all real numbers $y$, then the minimum value of $3y^2+24$ is 24.

Since the minimum value of $(x-3)^2 + 4(x-7)^2$ is 80 and the minimum value of $2(y-2)^2 + (y+4)^2$ is 24, then the minimum value of $(x-3)^2+2(y-2)^2 + 4(x-7)^2 + (y+4)^2$ is $80 + 24 = 104$.