# Homework 1995/01 - #5 Answers

## Problem 3

Using (Equation 77 - from Chapter 16) the patient's CL_{CR} = (140 - 56) x 79 / (72 x 2.0) = 46.1 ml/min
By plotting kel versus CL_{CR} it is possible to interpolate a value of kel for the patient

From this graph or by calculations the kel_{patient} = 0.0779 hr^{-1}

Proceeding a described on http://157.142.72.143../Ch16/Ch1606.html and first described on http://157.142.72.143../Ch14/Ch1408.html.

R = Cp_{min}/Cp_{max} = 1/6 = e^{-kel*tau}

Taking ln of both sides and solving for tau gives a value of 23.0 hr. Adjusting this **upwards** (to avoid extending the concentration range) give a new tau value of 24 hours.

The new R value is now e^{-0.0779 x 24} = 0.154

Maintenance dose can be calculated from Cp_{max} * V * (1 - R) = 6 x 0.28 x 79 x (1 - 0.154) = 112 mg (probably give 100 mg which would produce a Cp_{max} of 5.34 mg/L and Cp_{min} of 0.823 mg/L).

The required loading dose can be calculated as Cp_{max} * V = 6 x 0.28 x 79 = 133 mg (125 mg would produce a Cp_{max} of 5.65 mg/L).