Max p with p+q+r=5 & pq+qr+rp=3 (positives)
November 3, 2024 2025-11-03 23:36Max p with p+q+r=5 & pq+qr+rp=3 (positives)
Largest \(p\) given \(p+q+r=5\) and \(pq+qr+rp=3\) \((p,q,r>0)\)
Two symmetric constraints; by symmetry the maximum occurs at \(q=r\), and the rigorous check uses a discriminant. Final answer: \(\\dfrac{13}{3}\).
Problem
Positive reals \(p,q,r\) satisfy \[ p+q+r=5,\qquad pq+qr+rp=3. \] Find the largest possible value of \(p\).
One-line intuition (use symmetry)
Rigorous method (discriminant ≥ 0)
Full derivation
With \(p\) fixed, \(q\) and \(r\) are the roots of
\[
t^2-(5-p)t+(p^2-5p+3)=0.
\]
For real \(q,r\), the discriminant must be nonnegative:
\[
\Delta=(5-p)^2-4(p^2-5p+3)=-3p^2+10p+13\ge 0.
\]
This gives
\[
3p^2-10p-13\le 0 \;\Longrightarrow\; p\in\Big[-1,\ \frac{13}{3}\Big].
\]
Since \(p>0\), we get \(0 At \(p=\dfrac{13}{3}\): \(q+r=\dfrac{2}{3}\) and \(qr=\dfrac{1}{9}\) with \(\Delta=0\Rightarrow q=r=\dfrac{1}{3}\), which are positive. Hence the bound is attainable.
Quick recap
- From the sums: \(q+r=5-p\), \(qr=p^2-5p+3\).
- Feasibility needs discriminant \( \ge 0\Rightarrow p\le 13/3\).
- At the bound, \(q=r=1/3>0\). So the max is achieved.
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