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

2 Answers
Apr 7, 2018

The maximum value of Q is 2/3.

Explanation:

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

Apr 7, 2018

See below.

Explanation:

(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