$a + b + c + d = 1$ and $a^2 + b^2 + c^2 + d^2 = \frac{1}{3}$, where $-1 \le a,b,c,d \le 1$. Which value of $a$ is the largest possible?

Question

I'm a complete novice who honestly has no clue about how to solve these types of problems. This question was originally multiple choice, and I was able to get the answer $\frac12$ simply through process of elimination.
I would appreciate any help, pointing in the right direction.

Answer 1

The function to maximize is $f(a,b,c,d)=a$. The constraints are $g(a,b,c,d) = a+b+c+d=1$ and $h(a,b,c,d) = a^2+b^2+c^2+d^2 = \frac{1}{3}.$ So we solve the system

$$\nabla f = \lambda \nabla g +\mu \nabla h$$

which is

$$1 = \lambda + \mu 2a$$ $$0 = \lambda + \mu 2b$$ $$0 = \lambda + \mu 2c$$ $$0 = \lambda + \mu 2d$$

plus the two constraint equations. $6$ equations in $6$ variables. From the last 3 equations we see that $b=c=d.$ That makes the first constraint $a+3b=1$ and the second $a^2+3b^2 = \frac{1}{3}$ to get $a=0$ or $a=1/2$.

Answer 2

The Lagrange multiplier ansatz is powerful and quickly gives an answer, as demonstrated by B. Goddard: $\,a=\frac12$ and $\,b=c=d=\frac16\,$ does the job.

The given constraints may be connected by means of the Inequality between Arithmetic and Quadratic mean $$1-a \:=\: b+c+d \:\leqslant\: |b|+|c|+|d| \:\leqslant\: 3\sqrt{\frac{b^2+c^2+d^2}3} \:=\: \sqrt{1-3a^2}\,.$$ Squaring & rearranging yields $\:4a^2\leqslant2a\,,$ hence $\,a\,$ cannot be larger than $\frac12\,$.