What is the maximum value of #Q(p,q,r)=2pq+2pr+2qr# subject to #p+q+r=1#?

Question

2596 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-07T22:11:18

The maximum value of #Q# is #2/3#.

We want to maximise #Q(p,q,r)=2pr+2pq+2qr# subject to #p+q+r-1=0#.

Let #P(p,q,r)=q+p+r-1#. By the method of Lagrange multipliers, the extrema of #Q# occur where

#gradQ=lambdagradP#

#rArr((2q+2r),(2p+2r),(2p+2q))=lambda((1),(1),(1))#

So

#2q+2r=lambda# #(1)#

#2p+2r=lambda# #(2)#

#2p+2q=lambda# #(3)#

#(1)-(2)rArr2q-2p=0rArrp=q#

#(1)-(3)rArr2r-2p=0rArrp=r#

Since #p+q+r=1#, it follows that #p=q=r=1/3#

So the maximum value of #Q# is #2(1/3)(1/3)+2(1/3)(1/3)+2(1/3)(1/3)=2/3#

Answer 2 · 2018-04-07T22:20:21

See below.

#(p+q+r)^2 = p^2+q^2+r^2+Q(p,q,r) = 1# then

#1-Q(p,q,r) = p^2+q^2+r^2#

The solution is at the tangency points between the sphere

#S(p,q,r) = p^2+q^2+r^2#

with the plane

#Pi->p+q+r =1#

This point is for #p = q = r = 1/3 rArr Q(p,q,r) = 2/3#