How to maximize $p_1p_2$ subject to constraints?

Question

Given $x_2 \geq x_1 \geq 0$, solve the following optimization problem in $p_1$ and $p_2$.

$$\max p_1p_2$$

subject to:

$$p_1 x_1 + p_2 (x_2 - x_1) = 1 $$ $$0\leq p_2 \leq p_1$$

Answer 1

I had planned to answer your deleted question on maximum likelihood. I think the answer here depends on whether $x_2 \gt 2x_1$ or $x_2 \lt 2x_1$:

• If $x_2 \ge 2x_1$, I think the product is maximised when $p_1=\frac1{2x_1}$ and $p_2=\frac1{2(x_2-x_1)}$ which makes the product $p_1p_2 = \frac1{4x_1(x_2-x_1)}$, essentially for the reason Maria Mazur gives in her answer

• If $2x_1 \ge x_2 \ge x_1$, I think the product is maximised when $p_1=p_2=\frac1{x_2}$ which makes the product $p_1p_2 = \frac1{x_2^2}$

Note that, as you might expect, these are equal when $x_2 = 2x_1$ since they are then both $\frac1{4x_1^2} = \frac1{x_2^2}$

Answer 2

Hint

Substitute from the constraint in the object function (considering $0\le p_1\le p_2\le1$) and differentiate with respect to the remaining variable. Then compare it with the value of the object function in the limit points.

Answer 3

Let $p_2=t$, $x_2-x_1=y\geq 0$ and $x_1=x$, then $$p_1 = {1-ty\over x}$$ so $$p_1p_2 = t{1-ty\over x} = -{y\over x}t^2+{1\over x}t$$

If $y\neq 0$ then this quadratic equation achieves maximum at $t= {1\over 2y}$ and that maximum is $${1\over 4xy}={1\over 4x_1(x_2-x_1)}$$

If $y=0$ we get $$p_1p_2 = {1\over x}t\leq {1\over x}$$

so in this case maximum is $1/x$

Answer 4

Taking the lagrangian with the help of $\epsilon_1,\epsilon_2$ as slack variables,

$$ L(p,\lambda,\epsilon) = p_1p_2+\lambda_1(p_1 x_1+p_2(x_2-x_1)-1)+\lambda_2(p_2-\epsilon_1^2)+\lambda_3(p_1-p_2-\epsilon_2^2) $$

and solving the stationary conditions

$$ \nabla L = 0 $$

we get.

$$ \left[ \begin{array}{cccccccc} p_1 & p_2 & \lambda_1 & \lambda_2 & \lambda_3 & \epsilon_1^2 & \epsilon_2^2 & p_1 p_2\\ \frac{1}{x_1} & 0 & 0 & -\frac{1}{x_1} & 0 & 0 & \frac{1}{x_1} & 0 \\ \frac{1}{x_1} & 0 & 0 & -\frac{1}{x_1} & 0 & 0 & \frac{1}{x_1} & 0 \\ \frac{1}{2 x_1} & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2 x_1^2-2 x_1 x_2} & 0 & 0 & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & \frac 14\frac{1}{ x_1 x_2-x_1^2} \\ \frac{1}{2 x_1} & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2 x_1^2-2 x_1 x_2} & 0 & 0 & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & \frac 14\frac{1}{ x_1 x_2-x_1^2}\\ \frac{1}{2 x_1} & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2 x_1^2-2 x_1 x_2} & 0 & 0 & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & \frac 14\frac{1}{ x_1 x_2-x_1^2} \\ \frac{1}{2 x_1} & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2 x_1^2-2 x_1 x_2} & 0 & 0 & \frac{1}{2 \left(x_2-x_1\right)} & \frac{1}{2} \left(\frac{1}{x_1-x_2}+\frac{1}{x_1}\right) & \frac 14\frac{1}{ x_1 x_2-x_1^2} \\ \frac{1}{x_2} & \frac{1}{x_2} & -\frac{2}{x_2^2} & 0 & \frac{2 x_1-x_2}{x_2^2} & \frac{1}{x_2} & 0 & \frac{1}{x_2^2} \\ \frac{1}{x_2} & \frac{1}{x_2} & -\frac{2}{x_2^2} & 0 & \frac{2 x_1-x_2}{x_2^2} & \frac{1}{x_2} & 0 & \frac{1}{x_2^2} \\ \end{array} \right] $$

The number of solutions is due to the introduction of $\epsilon_i^2$. To be feasible one solution should have $\epsilon_i^2\ge 0$. Note also that the restriction is active when the corresponding $\epsilon_i$ is null.