Minimizing $a_1x_1^2 + a_2x_2^2$ for positive $a_i$, where $a_1x_1+a_2x_2=B$

Question

Find $$\min\{a_1x_1^2 + a_2x_2^2\}$$ Where $ a_1x_1 + a_2x_2 = B$, and $a_1>0$ and $a_2>0 $. Find $x_1$ and $x_2$.

Can we do it usig AG mean inequality?

Let's say we have the problem to find the minimum value of $ x_1^2 + x_2^2 $.

From: $ (x_1 - x_2)^2 ≥0 $, $ x_1^2 + x_2^2 ≥ 2x_1x_2 $

So the minimum value is: $2x_1x_2 $ for $x_1=x_2$.

Can be this done in a simillar manner for the starting problem.

why cannot we put $ x^2 =a_1 x_1^2, y^2 =a_2 x_2^2$ and solve it like root mean square inequality (without generalization?) We get: $x^2 = y^2 $
. $ a_1 x_1^2 = a_2 x_2^2 $

but not $x_1 = x_2$?

Answer 1

Going along your lines, we can write \begin{align} &a_1a_2(x_1-x_2)^2\ge0\\ \iff&a_1a_2(x_1^2+x_2^2)\ge 2a_1a_2x_1x_2\\ \iff&a_1^2x_1^2+a_2^2x_2^2+a_1a_2(x_1^2+x_2^2)\ge a_1^2x_1^2+a_2^2x_2^2+2a_1x_1a_2x_2\\ \iff&(a_1x_1^2+a_2x_2^2)(a_1+a_2)\ge(a_1x_1+a_2x_2)^2\\ \iff&a_1x_1^2+a_2x_2^2\ge\frac{B^2}{a_1+a_2} \end{align} This is the required lower bound.

Answer 2

Use standard algebra:

We have,

$$\begin{align}a_1x_1^2+a_2x_2^2&=a_1\left(\frac{B-a_2x_2}{a_1}\right)^2+a_2x_2^2\\ &=\frac{B^2-2Ba_2x_2+a_2^2x_2^2}{a_1}+a_2x_2^2\\ &=\left(a_2+\frac{a_2^2}{a_1}\right)x_2^2-\left(\frac{2Ba_2}{a_1}\right)x_2+\frac{B^2}{a_1}\end{align}$$

This is the quadratic polynomial. You can minimize the quadratic polynomial with the following method:

$$ax^2+bx+c=a(x-m)^2+n$$

where, $$m=-\frac{b}{2a}, ~ n=-\frac{b^2-4ac}{4a}$$

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In this case, we have

$$\begin{align}a:=a_2+\frac{a_2^2}{a_1}, ~b:=-\frac{2Ba_2}{a_1} , ~c:=\frac{B^2}{a_1}\end{align}$$

Hence we get,

$$m:=\frac{B}{a_1+a_2},~ n:=\frac{B^2}{a_1+a_2}$$

Thus,

$$\begin{align}a_1x_1^2+a_2x_2^2=\left(a_2+\frac{a_2^2}{a_1}\right)\left(x_2-\frac{B}{a_1+a_2}\right)^2+\frac{B^2}{a_1+a_2}\end{align}$$

Finally, we conclude that $$\min\left\{a_1x_1^2+a_2x_2^2 \mid a_1x_1+a_2x_2=B\right\}=\frac{B^2}{a_1+a_2}$$ which is attained at the point $x_1=x_2=B/(a_1+a_2)$.

Answer 3

I see there's nothing calculus-ish in the tags, so maybe a proof using calculus isn't allowed. If so then it's allowed here:

Lagrange multipliers. $f=a_1x_1^2+a_2x_2^2$, $g=a_1x_1+a_2x_2$, and we want $\nabla f=\lambda\nabla g$: $$\begin{align}2a_1x_1=\lambda a_1, \\2a_2x_2=\lambda a_2,\end{align}$$hence $x_1=x_2$, so $x_1=\dots$ and $x_2=\dots$.