Shoe production: full expenditure and a budget cap

Let $x,y,z$ count pairs of Walkers, Runners, Trainers. The printed solution assumes the entire budget is spent: $x+y+z=20$, $30x+40y+50z=980$, equivalently $3x+4y+5z=98$. The source GC parameterisation gives $x=z-18$, $y=38-2z$. With nonnegative integer counts, $18\le z\le19$, hence the two printed combinations $(x,y,z)=(0,2,18)$ and $(1,0,19)$.

The prompt says “a budget of” rather than explicitly requiring equality. Under the usual cap interpretation, $30x+40y+50z\le980$. All feasible combinations are $x=20-y-z$, with integers $0\le z\le19$ and $0\le y\le\min(20-z,38-2z)$. This gives 229 combinations, not just the two full expenditure cases. The printed equality assumption and its answers are retained above; this cap interpretation is an authored qualification. Zero counts are allowed, consistently with the source’s own two combinations.

Profit is $50x+90y+100z$. For the two source combinations, it is $0(80-30)+2(130-40)+18(150-50)=1980$ and $1(80-30)+0(130-40)+19(150-50)=1950$. Thus produce 2 pairs of Runners and 18 pairs of Trainers, for profit $1980, as printed. This is also uniquely optimal among all 229 budget cap combinations: with $x=20-y-z$, $P=1000+40y+50z$. For $z\le18$, maximise $y$ at $20-z$, giving $P\le1800+10z\le1980$; at $z=19$, the cap forces $y=0$, giving 1950.