You can
put this solution on YOUR website!dP/dx = -50x + 300
To find the max (or min), is where the gradient (dp/dx) is zero, so ask that...
0 = -50x + 300
hence x = 6.
Prove it is a max too, by differentiating again...d2P/dx2 = -50. This is -ve therefore MAX
So max number of clerks = 6.
This maximises the profit, P to be
max profit = 900
jon