The critical points is the point where the derivative of the density is 0 or does not exist. Therefore, the derivative of the density, ρ, with respect to temperature, T, is needed,
Since v is not 0, the critical point of density is the same as the volume and it is easy to calculate the critical point of the volume.
The formula of the volume is given
v = 999.87  0.06426T + 0.0085T^{2} 
0.00006T^{3}
Differentiate both side of the formula with respect to temperaure gives:
dv/dT =  0.06426 + 0.0170T  0.00018T^{2}
Find the temperature when dv/dT = 0.
 0.06426 + 0.0170T  0.00018T^{2} = 0
So T = 90.5 or T = 3.945
It is given that the range of temperature in the volume function is
from 0^{o}C to 40^{o}C, so the 90.5 is discarded and
thus 3.945 is the solution. At this temperature, the volume of the solution
is
v(3.945) = 999.87  0.06426T + 0.0085T^{2}

0.00006T^{3}
=
999.87  0.06426(3.945)
+
0.0085(3.945)^{2}  0.00006(3.945)^{3}
= 999.74
The density at temperature 3.945 is
ρ(3.945) = 1000/v = 1000/999.74 = 1.0003
With the same method, the density at the temperature of 0 and 40 can be calculated.
ρ(0) = 1.0001
ρ(3.945) = 0.993
Comparing the three density, the maximum value is 1.0003 kg/m^{3} at a temperature of 3.945^{o}C. 