Answer:
[tex]T \approx 3.967\,^{\textdegree}C[/tex]
Step-by-step explanation:
The density of water is given by the following definition:
[tex]\rho = \frac{m}{V(T)}[/tex]
[tex]\rho = \frac{1000\,g}{999.870.06426\cdot T + 0.0085043\cdot T^{2}-0.0000679\cdot T^{3}}[/tex]
The density is maximum when volume is minimum, which can be found by First and Second Derivative Tests:
First Derivative
[tex]V' = -0.06426 +0.0170086\cdot T -0.0002037\cdot T^{2}[/tex]
Second Derivative
[tex]V'' = 0.0170086 - 0.0004074\cdot T[/tex]
Critical values from the first derivative are:
[tex]T_{1} \approx 79.531\,^{\textdegree}C[/tex] (absolute maximum) and [tex]T_{2} \approx 3.967\,^{\textdegree}C[/tex] (absolute minimum).
The temperature at which water has its maximum density is:
[tex]T \approx 3.967\,^{\textdegree}C[/tex]
