How to minimize a 2D function given a constraint? How can I minimize
$$
2a^2 + b^2
$$
given the constraint that
$$
a+b = 1
$$
I am able to obtain the answer using Wolfram Alpha, but I'm not sure how to go through the math.

 A: Substitute using $b = 1-a$.
Then we have
$$
f(a) = 2a^2 + (1-a)^2 = 3a^2 - 2a + 1
$$
Then we have
$$
f^{\prime}(a) = 6a-2
$$
and setting to zero and solve we obtain $a=1/3$, so then because $a+b=1$, then $b=2/3$.
A: You could also use a giant hammer for the tiny nail here with Lagrange Multipliers: $F = 2a^2 + b^2 - \lambda(a+b -1)$ and minimize this with respect to $a$ and $b$ which will result in $F_a = 4a - \lambda = 0$ and $F_b = 2b - \lambda = 0$ thus $2a = b$ and using the constraint: $1 = a + b = 3a$ so $a = 1/3$ and the rest follows.
https://en.wikipedia.org/wiki/Lagrange_multiplier
A: Form an object function by combining with a Lagrange multiplier.
Or more easily solve together
$$2a^2+b^2= c^2, a+b=1$$
evaluate $(a,b) $ in terms of $c$ which yields
$$ 2a=b, a= \frac13,b= \frac23$$
A: Here it is an approach without calculus:
\begin{align*}
2a^{2} + b^{2} & = 2a^{2} + (1-a)^{2}\\\\
& = 3a^{2} -2a + 1\\\\
& = 3\left(a^{2} - \frac{2a}{3} + \frac{1}{9}\right) + \frac{2}{3}\\\\
& = 3\left(a-\frac{1}{3}\right)^{2} +\frac{2}{3} \geq \frac{2}{3}
\end{align*}
Hopefully this helps!
A: Here is a way using the well known Cauchy-Schwarz inequality (C.S.):
$$1^2 = \left((\sqrt{2}\cdot a)\frac 1{\sqrt 2} + b\cdot 1\right)^2\stackrel{C.S.}{\leq}(2a^2+b^2)\left(\frac 12 + 1\right)$$
Hence,
$$2a^2+b^2 \geq \frac 23$$
Equality is reached if $(\sqrt{2}\cdot a, b)$ is parallel to $\left(\frac 1{\sqrt 2},1\right)$, which happens for $a=\frac 13$ and $b= \frac 23$.
