Between 0^oC and 40^oC, the volume, v, of a 1 kg solution was determined to follow the relationship 

     v = 999.87 - 0.06426T + 0.0085T² - 0.00006T³

where T is the temperature. At what temperature is the mass density a maximum?

Let the density, volume and the mass of the solution be ρ, v and m respectively.

In order to find the maximum dencisty in the domain between 0^oC and 40^oC, the following steps need to be taken:

• Find the critical points of the function and calculate the value at critical points.
• Find the value of f(0) and f(40).
• Compare the result of the above steps. The maximum density is the largest one.

Critical Point for Density and
Volume are the Same

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 + 0.0085T² - 0.00006T³

Differentiate both side of the formula with respect to temperaure gives:

     dv/dT = - 0.06426 + 0.0170T - 0.00018T²

Find the temperature when dv/dT = 0.

     - 0.06426 + 0.0170T - 0.00018T² = 0

So T = 90.5 or T = 3.945

It is given that the range of temperature in the volume function is from 0^oC to 40^oC, 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 + 0.0085T²
                     - 0.00006T³

                  = 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/m³ at a temperature of 3.945^oC.