You want to minimize the function [tex]f(x,y)=x+3y[/tex] for [tex]x,y>0[/tex] subject to the constraint [tex]xy=1200[/tex].
By solving the constraint you get [tex]y=\frac{1200}{x}[/tex]. So you need to minimize the one-variable function [tex]f(x,\frac{1200}{x})=g(x)=x+\frac{3600}{x}[/tex].
Compute the derivative [tex]g'(x)=1-\frac{3600}{x^2}=\frac{x^2-3600}{x^2}[/tex]. The derivative is zero when [tex]x^2=3600[/tex]. You need [tex]x>0[/tex] so you get only one critical point [tex]x=60[/tex].
Since [tex]f'(1)<0[/tex] you get that [tex]f'(x)<0[/tex] for all [tex]x\in (0,60)[/tex] and since [tex]f'(61)>0[/tex] you get that [tex]f'(x)>0[/tex] for all [tex]x\in (60,+\infty)[/tex].
That means that [tex]g[/tex] has a minimum at [tex]x=60[/tex]. So the desired numbers are [tex]x=60[/tex] and [tex]y=\frac{1200}{60}=20[/tex].
