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# Between 0 $^\circ C$ and 30 $^\circ C$, the volume $V$ (in cubic centimeters) of $1$ kg of water at a temperature $T$ is given approximately by the formula$$V = 999.87 - 0.06426T + 0.0085043T^3 - 0.0000679T^3$$Find the temperature at which water has its maximum density.

## The density is defined as $\rho=\frac{\text { mass }}{\text { volume }}=\frac{1000}{V(T)}\left(\text { in } \mathrm{g} / \mathrm{cm}^{3}\right) .$ But a critical point of $\rho$ will also be a critical point of $V$$\left[\text { since } \frac{d \rho}{d T}=-1000 V^{-2} \frac{d V}{d T} \text { and } V \text { is never } 0\right], and V is easier to differentiate than \rho$V(T)=999.87-0.06426 T+0.0085043 T^{2}-0.0000679 T^{3} \Rightarrow V^{\prime}(T)=-0.06426+0.0170086 T-0.0002037 T^{2}$Setting this equal to 0 and using the quadratic formula to find T, we getT=\frac{-0.0170086 \pm \sqrt{0.0170086^{2}-4 \cdot 0.0002037 \cdot 0.06426}}{2(-0.0002037)} \approx 3.9665^{\circ} \mathrm{C} or 79.5318^{\circ} \mathrm{C}. since we are only interestedin the region 0^{\circ} \mathrm{C} \leq T \leq 30^{\circ} \mathrm{C}, we check the density \rho at the endpoints and at 3.9665^{\circ} \mathrm{C}: \rho(0) \approx \frac{1000}{999.87} \approx 1.00013$$\rho(30) \approx \frac{1000}{1003.7628} \approx 0.99625 ; \rho(3.9665) \approx \frac{1000}{999.7447} \approx 1.000255 .$ So water has its maximum density atabout $3.9665^{\circ} \mathrm{C}$

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

##### Michael J.

Idaho State University

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