Question

The extreme value of the function f(x1, x2,….. xn)=\(\frac{x_1}{2^0}+\frac{x_2}{2^1}+……+\frac{x_n}{2^{n-1}}\) With respect to the constraint Σ^mi=1 (xi)^2 = 1 where m always stays lesser than n and as m,n tends to infinity is?

(a) 1

(b) \(\frac{2}{3\sqrt{3}}\)

(c) 2

(d) 1 ⁄ 2

I got this question in an online interview.

Question is taken from Lagrange Method of Multiplier to Find Maxima or Minima in division Maxima and Minima of Engineering Mathematics

Answer 1

Correct answer is (b) \(\frac{2}{3\sqrt{3}}\)

To explain: First consider these functions as infinite dimension vectors. Given the constraint dimension is always less than the objective we can apply the Lagrange condition. We now have

\(\frac{1}{1}i_1+\frac{1}{2}i_2+\frac{1}{4}i_3+…..\infty=\lambda.(2x_1i_1+2x_2i_2+2x_3i_3+…\infty)\)

Treating equations individually we get

\(x_1=\frac{1}{2\lambda}\)

\(x_2=\frac{1}{4\lambda}\)

and so on.

Putting back in constraint we get

\(1=\frac{1}{(2\lambda)^2}+\frac{1}{(4\lambda)^2}+….\infty\)

\(1=\frac{1}{(\lambda)^2}+(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{8^2}….\infty)\)

\(\Rightarrow \lambda = \frac{1}{\sqrt{3}}\)

Hence, we get the extreme value (after putting back values of variables in the function) as

extreme value = \(\frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt[3]{3}}+\frac{1}{2\sqrt[5]{3}}+…\infty\)

extreme value = \(\frac{\frac{1}{2\sqrt{3}}}{1-\frac{1}{4}}=\frac{2}{3\sqrt{3}}\)