Brick orders: distinguish a fixed-profit system from cost minimisation

Let $p,q,r$ be the numbers of bricks ordered by P, Q, R. Then $p+q+r=30000$, $p+r=2q$ (equivalently $p-2q+r=0$), and $3r-3p=9600$ (equivalently $-p+r=3200$). The source GC solution is $p=8400,q=10000,r=11600$.

Discard the profit-difference equation as instructed; retain $p+q+r=30000$ and $p+r=2q$. Thus $q=10000$, $p=20000-r$. Minimum orders give $1000\le r\le19000$. Transport cost is $\begin{aligned}&C\\ & =0.04p+0.02(10000)\\ &{}+0.03r\\ & =0.04(20000-r)+200\\ &{}+0.03r\\ & =1000-0.01r.\end{aligned}$ It decreases with $r$, so choose $r=19000$, $p=1000$, $q=10000$ as printed. The minimum cost is $810 (an authored check). The source’s intermediate expansion writes 2000 instead of 200, but its next expression $1000-0.01r$ uses the correct 200.