Impedance Matching

(* source impedance = 1, load impedance = x *)
loadvoltage[x_] = x/(x+1)
Plot[loadvoltage[x],{x,0,10}]

loadpower[x_] = x/(x+1)^2
Plot[loadpower[x],{x,0,10}]

Note the maximum load power at x = 1, when load impedance matches source impedance.

D[loadpower[x],x]

Solve[%==0,x]