Let #r#, #s#, and #t# be the roots of the equation #x^3 - 2x + 1 = 0# in some order. What is the maximal value of #r^3 - s- t#?

Please answer!

1 Answer
May 8, 2017

#1/2 (5 + 3 sqrt[5])#

Explanation:

This optimization problem can be solved with the contribution of the so called Lagrange Multipliers

Calling

#f(r,s,t) = r^3-s-t#

and the restrictions

#g_1(r,s,t)=r+s+t=0#
#g_2(r,s,t)=rs+st+rt+2=0#
#g_3(r,s,t)=rst-1=0#

and forming the lagrangian

#L(r,s,t,lambda_1,lambda_2,lambda_3)=f(r,s,t)+lambda_1g_1(r,s,t)+lambda_2g_2(r,s,t)+lambda_3g_3(r,s,t)#

The stationary points are found by solving

#grad L = 0# or

#{(lambda_1 + 3 r^2 + lambda_2 s t + lambda_3 (s + t)=0),( lambda_1 + lambda_2 r t + lambda_3 (r + t)-1=0),(lambda_1 + lambda_2 r s + lambda_3 (r + s)-1),(r + s + t=0),( r s + r t + s t+2=0),(r s t-1=0):}#

and the solutions for maximum value are

#((r,s,t,lambda_1,lambda_2,lambda_3),(1/2 (1 + sqrt[5]), -1, 1/2 (1 - sqrt[5]),1/10 (-5 - 11 sqrt[5]), 1/10 (5 - 9 sqrt[5]), 2 + 1/sqrt[5]),(1/2 (1 + sqrt[5]),1/2 (1 - sqrt[5]), -1, 1/10 (-5 - 11 sqrt[5]),1/10 (5 - 9 sqrt[5]), 2 + 1/sqrt[5]))#

giving the maximum as

#1/2 (5 + 3 sqrt[5])#

NOTE:
This is an almost automatic procedure to obtain the maximum. A shorter way would be extracting the roots, which is easy and then making root arrangements and computing for #f(r,s,t)#.