Let n ? 2. If xi is a Boolean variable for all 1 ? i ? n, prove that a) (x1 + x2
ID: 1892075 • Letter: L
Question
Let n ? 2. If xi is a Boolean variable for all 1 ? i ? n, prove thata) (x1 + x2 + ... + xn) x1x2 ... xn
b) (x1x2 ... xn) x1 + x2 + ... + xn
Explanation / Answer
We use Lagrange multipliers, where g(x1; : : : ; xn) = x1 + : : : + xn. The gradients of f; g are rf = x2 : : : xn n(x1 : : : xn) 11=n ; : : : ; x1 : : : xn1 n(x1 : : : xn) 11=n ; rg = h1; : : : ; 1i: Therefore the Lagrange multiplier system reads = x2 : : : xn n(x1 : : : xn) 11=n ; : : : ; x1 : : : xn1 n(x1 : : : xn) 11=n : Notice that the rst term equation can be rewritten as = pn x1 : : : xn 1 nx1 : We can do something similar for every other equation, and we obtain n equa- tions = pn x1 : : : xn 1 nxi : Remembering that all the xi > 0, we can set all these terms on the right of these equations equal to each other (as they all equal ), and then divide by the common factor of pn x1 : : : xn=n to get x1 = x2 = : : : = xn. Since g(x1; : : : ; xn) = x1 + : : : + xn = c, the solution to the Lagrange multiplier system consists of the points (c=n; c=n; : : : ; c=n), and the value of f at this point is pn (c=n) n = c=n. We now need to check that this is an absolute maximum. Since the con- straint region is not of the form g(x; y) = c, but rather of the form g(x; y) = c together with xi > 0, we need to check what happens near the boundary of the constraint region. If we are near the constraint region, one of the xi , say x1, is very close to 0. On the other hand, all the remaining xj are bounded from above by c, so as we approach the boundary of the constraint region, f(x1; : : : xn) = pn x1 : : : xn x 1=n 1 c ! 0 as x1 ! 0. Therefore the point (c=n; : : : ; c=n) is an absolute maximum for f on our constraint region.
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