# quadratic functions to find $a$

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.

• – John Omielan May 4 at 8:46

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 $$x=-\frac{b}{2a}$$