Quadratic Forms and Objective functions with two or more variables

Two Choice Variables Quadratic Forms Definiteness Let 𝑢=𝑑𝑥, 𝑣=𝑑𝑦, 𝑎= 𝑓 𝑥𝑥 , 𝑏= 𝑓 𝑦𝑦 , ℎ= 𝑓 𝑥𝑦 = 𝑓 𝑦𝑥 Then 𝑑 2 𝑍= 𝑓 𝑥𝑥 𝑑 𝑥 2 +2 𝑓 𝑥𝑦 𝑑𝑦𝑑𝑥+ 𝑓 𝑦𝑦 𝑑 𝑦 2 ------------- (𝑒𝑞𝑛 1) can be written as 𝐝 𝟐 𝐙=𝒒=𝒂 𝒖 𝟐 +𝟐𝒉𝒖𝒗+𝒃 𝒗 𝟐 ------------- ---- (𝑒𝑞𝑛 2)  Definiteness Positive definite if q is invariably positive (q > 0) Positive semidefinite if q is invariably nonnegative (𝑞≥0 ) Negative definite if q is invariably negative (q < 0) Negative semidefinite if q is invariably non positive (𝑞≤0 )

Rewriting the quadratic form using completing square 𝒂 𝒖+ 𝒉 𝒂 𝒗 𝟐 +𝒂𝒃− 𝒉 𝟐 𝒂 𝒗 𝟐 𝑞 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑖𝑓𝑓 𝑎>0 𝑎𝑛𝑑 𝑎𝑏− ℎ 2 >0 𝑞 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑖𝑓 𝑎<0 𝑎𝑛𝑑 𝑎𝑏− ℎ 2 >0 Writing in Matrix Form We have 𝑢 𝑣 𝑎 ℎ ℎ 𝑏 𝑢 𝑣   The determinant of the coefficient matrix 𝑎 ℎ ℎ 𝑏 is important in determining the sign of 𝑞 𝑞 𝑖𝑠 𝒑𝒐𝒔𝒊𝒕𝒊𝒗𝒆 𝒅𝒆𝒇𝒊𝒏𝒊𝒕𝒆 𝑖𝑓𝑓 ∣𝑎∣>0 𝑎𝑛𝑑 𝑎 ℎ ℎ 𝑏 >0 𝑞 𝑖𝑠 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 𝒅𝒆𝒇𝒊𝒏𝒊𝒕𝒆 𝑖𝑓𝑓 ∣𝑎∣<0 𝑎𝑛𝑑 𝑎 ℎ ℎ 𝑏 >0

Writing equation (2) in terms of equation (1) gives further insights Since 𝑎= 𝑓 𝑥𝑥 𝑎𝑛𝑑 𝑎𝑏− ℎ 2 𝑖𝑠 𝑓 𝑥𝑥 𝑓 𝑦𝑦 − 𝑓 𝑥𝑦 . 𝑓 𝑦𝑥 𝑓 𝑥𝑥 𝑓 𝑦𝑦 − 𝑓 𝑥𝑦 . 𝑓 𝑦𝑥 𝑜𝑟 ( 𝑓 𝑥𝑥 𝑓 𝑦𝑦 − 𝑓 𝑥𝑦 2 ) Conditions for 𝑑 2 𝑍 𝑓𝑜𝑟 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑖𝑠 𝑓 𝑥𝑥 > 0 𝑎𝑛𝑑 𝑓 𝑥𝑥 𝑓 𝑥𝑦 𝑓 𝑥𝑦 𝑓 𝑦𝑦 >0 𝑑 2 𝑍 𝑓𝑜𝑟 𝑛𝑒𝑔𝑖𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑖𝑠 𝑓 𝑥𝑥 <0 𝑎𝑛𝑑 𝑓 𝑥𝑥 𝑓 𝑥𝑦 𝑓 𝑥𝑦 𝑓 𝑦𝑦 >0

Discriminant of a Quadratic form In general the discriminant of a quadratic form 𝑞=𝑎 𝑢 2 +2ℎ𝑢𝑣+𝑏 𝑣 2 Is the symmetric determinant 𝑎 ℎ ℎ 𝑏 In the particular case of quadratic form 𝑑 2 𝑍= 𝑓 𝑥𝑥 𝑑 𝑥 2 +2 𝑓 𝑥𝑦 𝑑𝑦𝑑𝑥+ 𝑓 𝑦𝑦 𝑑 𝑦 2 The discriminant is the determinant with second order partial derivatives as its elements.

Hessian Determinant ∣𝐻∣= 𝒇 𝒙𝒙 𝒇 𝒙𝒚 𝒇 𝒚𝒙 𝒇 𝒚𝒚 Determinant with all the second order partial derivatives is called Hessian Matrix.

Extremum Conditions for two choice variable 𝑖𝑓 𝑑 2 𝑍 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑖𝑡 𝑖𝑠 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 𝑖𝑓 𝑑 2 𝑍 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑡ℎ𝑒𝑛 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 So, 𝑑 2 𝑍 𝑓𝑜𝑟 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒(𝑚𝑖𝑛𝑖𝑚𝑢𝑚) 𝑖𝑓 𝑓 𝑥𝑥 > 0 𝑎𝑛𝑑 𝑓 𝑥𝑥 𝑓 𝑥𝑦 𝑓 𝑥𝑦 𝑓 𝑦𝑦 >0 𝑑 2 𝑍 𝑓𝑜𝑟 𝑛𝑒𝑔𝑖𝑡𝑖𝑣𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑖𝑓 𝑓 𝑥𝑥 < 0 𝑎𝑛𝑑 𝑓 𝑥𝑥 𝑓 𝑥𝑦 𝑓 𝑥𝑦 𝑓 𝑦𝑦 >0  

Objective Functions with 3 choice Variables 𝑧=𝑓 𝑥 1 , 𝑥 2 , 𝑥 3 𝑑𝑧= 𝑓 1 𝑑 𝑥 1 + 𝑓 2 𝑑 𝑥 2 + 𝑓 3 𝑑 𝑥 3 First order condition: 𝑑𝑧=0, 𝑖.𝑒. 𝑓 1 = 𝑓 2 = 𝑓 3 =0 Second order condition : 𝑑 2 𝑍 𝑑 2 𝑍= 𝜕 𝜕 𝑥 1 𝑓 1 𝑑 𝑥 1 + 𝑓 2 𝑑 𝑥 2 + 𝑓 3 𝑑 𝑥 3 𝑑 𝑥 1 + 𝜕 𝜕 𝑥 2 𝑓 1 𝑑 𝑥 1 + 𝑓 2 𝑑 𝑥 2 + 𝑓 3 𝑑 𝑥 3 𝑓 1 𝑑 𝑥 1 + 𝑓 2 𝑑 𝑥 2 + 𝑓 3 𝑑 𝑥 3 𝑑 𝑥 2 + 𝜕 𝜕 𝑥 3 𝑓 1 𝑑 𝑥 1 + 𝑓 2 𝑑 𝑥 2 + 𝑓 3 𝑑 𝑥 3 𝑑 𝑥 3 = 𝑓 11 𝑑 𝑥 1 2 + 𝑓 12 𝑑 𝑥 1 𝑑 𝑥 2 + 𝑓 13 𝑑 𝑥 1 𝑑 𝑥 3 + 𝑓 21 𝑑 𝑥 2 𝑑 𝑥 1 + 𝑓 22 𝑑 𝑥 2 2 + 𝑓 23 𝑑 𝑥 2 𝑥 3 + 𝑓 31 𝑑 𝑥 3 𝑑 𝑥 1 + 𝑓 32 𝑑 𝑥 3 𝑑 𝑥 2 + 𝑓 33 𝑑 𝑥 3 2

Hessian Matrix for 3 choice Variables 𝐻= 𝑓 11 𝑓 12 𝑓 13 𝑓 21 𝑓 22 𝑓 23 𝑓 31 𝑓 32 𝑓 33 𝑤ℎ𝑒𝑟𝑒 ∣ 𝐻 1 ∣= 𝑓 11 , ∣ 𝐻 2 ∣= 𝑓 11 𝑓 12 𝑓 21 𝑓 22 𝑎𝑛𝑑 ∣ 𝐻 3 ∣= 𝑓 11 𝑓 12 𝑓 13 𝑓 21 𝑓 22 𝑓 23 𝑓 31 𝑓 32 𝑓 33 Optimization Conditions First Order Necessary condition: 𝒇 𝟏 = 𝒇 𝟐 = 𝒇 𝟑 =𝟎 Second order Conditions 𝑓𝑜𝑟 𝑍 ∗ 𝑡𝑜 𝑏𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 ∣ 𝐻 1 ∣<0, ∣ 𝐻 2 ∣>0, ∣ 𝐻 3 ∣<0 ; 𝑑 2 𝑍(𝑁𝐷) 𝑓𝑜𝑟 𝑍 ∗ 𝑡𝑜 𝑏𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 ∣ 𝐻 1 ∣>0, ∣ 𝐻 2 ∣>0, ∣ 𝐻 3 ∣>0 ; 𝑑 2 𝑍(𝑃𝐷)

𝑵 Variable Case First order Necessary conditions Optimization Conditions First order Necessary conditions 𝑓 1 = 𝑓 2 = 𝑓 3 = 𝑓 4 =…………… 𝑓 𝑛 =0 Second order Conditions 𝑓𝑜𝑟 𝑍 ∗ 𝑡𝑜 𝑏𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 ∣ 𝐻 1 ∣<0, ∣ 𝐻 2 ∣>0, ∣ 𝐻 3 ∣< 0,∣ 𝐻 4 ∣>0… −1 𝑛 ∣ 𝐻 𝑛 ∣>0 ; 𝑑 2 𝑍(𝑁𝐷) 𝑓𝑜𝑟 𝑍 ∗ 𝑡𝑜 𝑏𝑒 𝑚𝑖𝑛𝑖𝑚𝑢𝑚 ∣ 𝐻 1 ∣>0, ∣ 𝐻 2 ∣>0, ∣ 𝐻 3 ∣> 0 ….∣ 𝐻 𝑛 ∣>0; 𝑑 2 𝑍(𝑃𝐷)