If we have objective function involving n variables:

U = f(X₁,. X₂. X₃ ..... X_n) subject to ϕ(X1. X2. X3 ..... Xn) = 0. The second order condition for extremum to be

(A) For maximum \(|\overline{H}_2|<0.|\overline{H}_3|<0....\)

(B) For maximum \(|\overline{H}_2|>0.|\overline{H}_3|<0....\)

(C) For maximum \(|\overline{H}_2|<0.|\overline{H}_3|>0....\)

(D) For minimum \(|\overline{H}_2|<0.|\overline{H}_3|<0....\)

(E) For minimum \(|\overline{H}_2|>0.|\overline{H}_3|>0....\)

Choose the correct answer from the options given below:

Question

Question

This question was previously asked in

UGC NET Paper 2: Economics 8 Oct 2022 Shift 2

Answer 1

The correct answer is (B) and (D) onlyKey Points

When there're n independent variables, the objective function may be expressed as

U = f{x\, X2, •... x-).

du = f₁ dx₁ + --- + f_ndx_n

So that the first order condition for minimum or maximum will be that all the n first order partial derivatives must vanish (since du = 0) and therefore:

f₁ = f₂ ---- =f_n = 0.

For the second order condition
- d²z will have a positive sign
- if all the n principal minors of the Hessian determinant: \(|H| = \begin{bmatrix} f_{11} & {--} & f_{1n} \\[.3em] f_{21} & f_{22} & f_{2n} \\[0.3em] -- & -- & --\\[0.3em] f_{n1} &f_{n2}&f_{nn} \end{bmatrix} \text{are positive}\)
- d²u will have a negative sign, functi~n will have maximum value if n
  principal minors of H possess alternate signs - the first being negative.
Hence For u = f(x1,x2,--,xa) to be