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If given $ a+b=1$, Var$(x̄)=9$ and Var$(ȳ)=16$, then Var$(ax̄+bȳ)=9a^2+16b^2=9a^2+16(1-a)^2=25a^2-32a+16$. Why the next step is ${}=(5a-\frac{16}{5})^2+16-(\frac{16}{5})^2$ and from this, why we can conclude that when $a=\frac{16}{25}$

$$\mathbb{V}(ax̄+bȳ)$$

gets minimum?
I don't really understand how to generate the next step and find the a value from it.

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as you calculated,

$$\mathbb{V}[a\overline{X}+b\overline{Y}]=25a^2-32a+16$$

which is a parabola equation, with minimum in

$$a=\frac{32}{50}=\frac{16}{25}$$


remember that in the parabola equation

$$y=ax^2+bx+c$$

its minimum is attained when $x=-\frac{b}{2a}$

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