# Minimum and maximum

There are five real numbers $a,b,c,d,e$ such that

$$a + b + c + d + e = 7$$$$a^2+b^2+c^2+d^2+e^2 = 10$$

How can we find the maximum and minimum possible values of any one of the numbers ?

Source

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Maximum value that can be taken by one of the variables is $\frac{9}{5}$ and the minimum value is $1$.

This can be argued based on symmetry as follows (and I hope I have not overlooked anything in my symmetry argument).

Say we want to find the extreme values of $a$. By symmetry when $a$ reaches the extremum the remaining variables have to be equal. So when $a$ takes the extremum $b=c=d=e$.

Hence, $a + 4b = 7$ and $a^2 + 4b^2 = 10$ i.e. $a^2 + 4 (\frac{7-a}{4})^2 = 10$

i.e. $a^2 + \frac{(a-7)^2}{4} = 10$ i.e. $5a^2 -14a + 49 = 40$ i.e. $5a^2 - 14a + 9 =0$.

Hence, $(5a-9)(a-1)=0$.

So, minimum is $a=1$ and maximum is $a = \frac{9}{5}$.

The round about way would be to use Lagrange multiplier technique and set up 7 equations in 7 unknowns and solve it.

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Actually, if you don't trust the symmetry argument (I think it's correct but needs a bit of justification), you can mix the two approaches -- if you write down the objective function with Lagrange multipliers and differentiate it with respect to one of the variables you're not optimizing, you see immediately that you'll get the same linear equation for each of them, so you know they'll come out equal and can switch to the symmetry approach with a peaceful mind :-) –  joriki Feb 8 '11 at 19:33
@joriki: True. The equations obtained from Lagrange multiplier will turn out to be symmetric in the remaining variables. –  user17762 Feb 8 '11 at 20:01

I just wanted to point out that if you don't want to appeal to symmetry there is a Lagrange multiplier argument which isn't too bad: Plugging in the first equation into the second, the last 4 variables satisfy the equation $$(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2 = 10$$ Suppose you want to maximize $b$ for example. Then the gradient of $(7 - b - c - d - e)^2 + b^2 + c^2 + d^2 + e^2$ has to be a multiple of the gradient of $b$, which means there is some $\lambda$ such that $$-2(7 - b - c - d - e) + 2b = \lambda$$ $$-2(7 - b - c - d - e) + 2c = 0$$ $$-2(7 - b - c - d - e) + 2d = 0$$ $$-2(7 - b - c - d - e) + 2e = 0$$ But the last three equations are the same as $-2a + 2c = 0$, $-2a + 2d = 0$, and $-2a + 2e = 0$, which lead to $a = c = d = e$. Thus the $4$ variables other than $b$ are the same and one can now proceed as in Sivaram's argument.

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So you want $x\in \mathbb{R}^5$ such that $\| x\|^2 = 10$ and $(x,1)=7$. (Here 1 refers to the all 1's vector)

Geometrically, we are looking at a circle around the all 1's vector in $\mathbb{R}^5$, and we just need to find the extremes of the circle.

From the above discussion it should be clear that the maximum and minimum will be close to $\sqrt {2}$, but the question is how much can it deviate?

Now lets actually solve the problem: Consider $a=b=c=d$ so our equations are in two variables, $$4a+e=7$$
$$4a^2+e^2=10$$

Solving $a=\frac{7-e}{4}$ and plugging into the second equation we get the quadratic equation: $$5e^2-14e+9=0$$ Which factors nicely as $$(5e-9)(e-1)=0$$ So either $e=1$ or $e=1.8$. I now claim that these are indeed the minimum and maximum values, but I will leave that for you to prove.

Hint: The vector $x$ in $\mathbb {R}^n$ with $\|x\|\leq r$ that maximizes $(x,1)$ will be in the direction of $1$ (the all 1's vector). In other words, maximizing $a+b+c+d$ given $a^2+b^2+c^2+d^2<10-1.8^2$ occurs when $a=b=c=d$. Similarly for minimizing.

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