Answer
The temperature at which water has its maximum density is $3.97^{\circ}C$
Work Step by Step
We can write an expression for the density $\rho$:
$\rho = \frac{mass}{volume}$
Then, to find the maximum density for a given mass, we need to find the minimum volume.
We can find the value of $T$ when $V' = 0$:
$V = 999.87 - 0.06426T + 0.0085043T^2 - 0.0000679T^3$
$V' = -0.06426 + (2)(0.0085043T) - (3)(0.0000679T^2) = 0$
$0.0002037T^2-0.0170086T +0.06426 = 0$
We can use the quadratic formula to find the values of $T$:
$T = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$T = \frac{-(-0.0170086) \pm \sqrt{(-0.0170086)^2-4(0.0002037)(0.06426)}}{(2)(0.0002037)}$
$T = \frac{0.0170086 \pm \sqrt{(-0.0170086)^2-4(0.0002037)(0.06426)}}{0.0004074}$
$T = 3.97~~$ or $T = 79.5$
Since the domain is between $0^{\circ}C$ and $30^{\circ}C$, the solution is $T = 3.97^{\circ}C$
We can verify the volume for $T = 3.97^{\circ}C$ and the endpoints of the interval $T = 0^{\circ}C$ and $T = 30^{\circ}C$:
$V(3.97) = 999.87 - 0.06426(3.97) + 0.0085043(3.97)^2 - 0.0000679(3.97)^3 = 999.74~cm^3$
$V(0) = 999.87 - 0.06426(0) + 0.0085043(0)^2 - 0.0000679(0)^3 = 999.87~cm^3$
$V(30) = 999.87 - 0.06426(30) + 0.0085043(30)^2 - 0.0000679(30)^3 = 1003.76~cm^3$
The volume is minimized when $T = 3.97^{\circ}C$, which maximizes the density.
The temperature at which water has its maximum density is $3.97^{\circ}C$
