Carly Cannon

Answered

2022-07-02

I have the following equations and inequalities:

$1={A}^{\prime}+{B}^{\prime}$

$1=A+B+C$

$A\le {A}^{\prime}$

$B\le {B}^{\prime}$

All variables are bounded below by zero and above by one.

I wonder if I can find an analytic expression for the upper and lower bounds for the difference ${A}^{\prime}(1-C)-A$

$1={A}^{\prime}+{B}^{\prime}$

$1=A+B+C$

$A\le {A}^{\prime}$

$B\le {B}^{\prime}$

All variables are bounded below by zero and above by one.

I wonder if I can find an analytic expression for the upper and lower bounds for the difference ${A}^{\prime}(1-C)-A$

Answer & Explanation

lofoptiformfp

Expert

2022-07-03Added 16 answers

Step 1

First note ${A}^{\prime}(1-C)-A={A}^{\prime}(A+B)-A$

For the upper bound clearly we need to take $A=0$ , $B={B}^{\prime}$ and $C={A}^{\prime}$ . In which case the problem comes down to finding the maximum value of ${A}^{\prime}{B}^{\prime}={A}^{\prime}-{A}^{\prime 2}$ which is obtained if ${A}^{\prime}=\frac{1}{2}$ so an upper bound is $\frac{1}{4}$. For a lower bound clearly we need to take $A={A}^{\prime}$ , $B=0$ and $C={B}^{\prime}$ , so we have to find the minimum value of ${A}^{2}-A$ for $0\le A\le 1$ , which is obtained if $A=\frac{1}{2}$. So a lower bound is $-\frac{1}{4}$ .

So $-\frac{1}{4}\le {A}^{\prime}(1-C)-A\le \frac{1}{4}$ .

First note ${A}^{\prime}(1-C)-A={A}^{\prime}(A+B)-A$

For the upper bound clearly we need to take $A=0$ , $B={B}^{\prime}$ and $C={A}^{\prime}$ . In which case the problem comes down to finding the maximum value of ${A}^{\prime}{B}^{\prime}={A}^{\prime}-{A}^{\prime 2}$ which is obtained if ${A}^{\prime}=\frac{1}{2}$ so an upper bound is $\frac{1}{4}$. For a lower bound clearly we need to take $A={A}^{\prime}$ , $B=0$ and $C={B}^{\prime}$ , so we have to find the minimum value of ${A}^{2}-A$ for $0\le A\le 1$ , which is obtained if $A=\frac{1}{2}$. So a lower bound is $-\frac{1}{4}$ .

So $-\frac{1}{4}\le {A}^{\prime}(1-C)-A\le \frac{1}{4}$ .

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