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

Apr 7, 2018

#### Answer:

The maximum value of $Q$ is $\frac{2}{3}$.

#### Explanation:

We want to maximise $Q \left(p , q , r\right) = 2 p r + 2 p q + 2 q r$ subject to $p + q + r - 1 = 0$.

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

$\nabla Q = \lambda \nabla P$

$\Rightarrow \left(\begin{matrix}2 q + 2 r \\ 2 p + 2 r \\ 2 p + 2 q\end{matrix}\right) = \lambda \left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)$

So

$2 q + 2 r = \lambda$ $\left(1\right)$

$2 p + 2 r = \lambda$ $\left(2\right)$

$2 p + 2 q = \lambda$ $\left(3\right)$

$\left(1\right) - \left(2\right) \Rightarrow 2 q - 2 p = 0 \Rightarrow p = q$

$\left(1\right) - \left(3\right) \Rightarrow 2 r - 2 p = 0 \Rightarrow p = r$

Since $p + q + r = 1$, it follows that $p = q = r = \frac{1}{3}$

So the maximum value of $Q$ is $2 \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) + 2 \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) + 2 \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) = \frac{2}{3}$

Apr 7, 2018

See below.

#### Explanation:

${\left(p + q + r\right)}^{2} = {p}^{2} + {q}^{2} + {r}^{2} + Q \left(p , q , r\right) = 1$ then

$1 - Q \left(p , q , r\right) = {p}^{2} + {q}^{2} + {r}^{2}$

The solution is at the tangency points between the sphere

$S \left(p , q , r\right) = {p}^{2} + {q}^{2} + {r}^{2}$

with the plane

$\Pi \to p + q + r = 1$

This point is for $p = q = r = \frac{1}{3} \Rightarrow Q \left(p , q , r\right) = \frac{2}{3}$