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Let $a,b,$ be real numbers such that $a\ge 4b$ . Find the minimum of $(a-b)^2+(a-1)^2+(b-1)^2$.

This problem is http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=553680

River Li
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math110
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4 Answers4

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The minimum is either on the interior of the feasible set, or on the boundary.

The gradient of the objective is $$\left[2(a-b)+2(a-1), -2(a-b)+2(b-1)\right] = \left[4a-2b-2, 4b-2a-2\right],$$ and so the unconstrained minimum is the infeasible point $(1,1)$. Therefore the minimum is on the boundary. Substituting $a=4b$ gives $$\min_b\ (3b)^2 + (4b-1)^2 + (b-1)^2 = \min_b\ 26b^2-10b+2.$$ The gradient is $52b-10$, so $b=\frac{5}{26}$ and the minimum is $\frac{27}{26}$.

user7530
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  • "The minimum is either on the interior of the feasible set, or on the boundary." How do you justify this? – TonyK Sep 12 '13 at 17:56
  • Given sufficient regularity conditions (which are trivially satisfied in this problem) the KKT conditions hold: at every critical point $0 = \nabla f(x) + \lambda \nabla g(x)$, where $f$ is the objective and $g$ the constraint, with $\lambda g = 0$. If $\lambda=0$ the minimum is also an unconstrained minimum $0 = \nabla f$. If not, $g=0$ and you are on the boundary. – user7530 Sep 12 '13 at 19:03
  • I see that I phrased my question badly. Obviously the minimum value of the objective function in the feasible set is either in its interior or on its boundary! What I meant was: Just because we know that the global minimum is outside the feasible set, why does that imply that the minimum in the feasible set is on its boundary? And I see that it is because the objective function has just one local minimum (which lies outside the feasible set). Unless I have misunderstood you, I don't think regularity conditions are sufficient. – TonyK Sep 13 '13 at 17:05
  • They're not, but the objective is convex so there is only one unconstrained critical point. – user7530 Sep 13 '13 at 17:16
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Hint: Let $a=4b+c^2$. You can minimize the resulting function of $b$ and $c$, without worrying about a boundary.

Barry Cipra
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Set $a = 4b + c\ $ and then minimize the function $f(b,c) = (4b +c -b)^2 + (4b+c-1)^2 + (b-1)^2$ using the standard derivative test. The function will be at a minimum on the half-plane where it is defined either a local minimum, or along the boundary, which is where $c=0$.

Daron
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Remarks: Here is a solution without calculus. Similar trick is used in e.g. this.

We have the following identity \begin{align*} &(a - b)^2 + (a - 1)^2 + (b - 1)^2\\ ={}& \frac{27}{26} + \frac{9}{13}(a - 4b) + 2\left(a - \frac{b}{2} - \frac{35}{52}\right)^2 + \frac{3}{2}\left(b - \frac{5}{26}\right)^2. \end{align*}

Thus, we have $(a - b)^2 + (a - 1)^2 + (b - 1)^2 \ge \frac{27}{26}$ for all real numbers $a\ge 4b$.

Also, when $a = 10/13, b = 5/26$, we have $(a - b)^2 + (a - 1)^2 + (b - 1)^2 = 27/26$.

Thus, the minimum of $(a - b)^2 + (a - 1)^2 + (b - 1)^2$ subject to $a \ge 4b$ is $\frac{27}{26}$ when $b = \frac{5}{26}$ and $a = \frac{10}{13}$ (satisfying $a = 4b$).

$\phantom{2}$

Remarks: How to obtain this identity?

We use the method of unknown coefficients.

We hope to find constants $M$ and $\alpha \ge 0$ such that $$f(a, b) := (a - b)^2 + (a - 1)^2 + (b - 1)^2 - M - \lambda (a - 4b) $$ is non-negative for all $a, b \in \mathbb{R}$. If so, we have $(a-b)^2+(a-1)^2 + (b-1)^2 \ge M$ for reals $a\ge 4b$. Furthermore, when $a = 4b$, we have $(a-b)^2+(a-1)^2 + (b-1)^2 \ge M$.

We complete the square in two variables ($a$ and $b$), one at a time. We have $$f(a, b) = 2\left(a - \frac{\lambda}{4} - \frac{b}{2} - \frac12\right)^2 + \frac32\left(b - 1 + \frac76\lambda\right)^2 - M - \frac{13}{6}\lambda^2 + 3\lambda.$$

Solving the system of \begin{align*} a - \frac{\lambda}{4} - \frac{b}{2} - \frac12 &= 0, \\ b - 1 + \frac76\lambda &= 0, \\ - M - \frac{13}{6}\lambda^2 + 3\lambda &= 0,\\ a - 4b &= 0. \end{align*} we have $\lambda = \frac{9}{13}$ and $M = \frac{27}{26}$.

River Li
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