1. A flower vase in the form of a hexagonal prism, is to be filled with 512 cu.in of water. Find the height of the water if the wet portion of the flower vase and its volume are numerically equal.
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Volume:
$V = A_b h$
$512 = \left[ 6 \times \frac{1}{2}(x^2)\sin 60^\circ \right] h$
$\frac{3}{2}\sqrt{3}x^2 h = 512$
$h = \dfrac{1024}{3\sqrt{3} \, x^2}$
Wet area:
$A = A_b + A_L$
$512 = 6 \times \frac{1}{2}(x^2)\sin 60^\circ + 6xh$
$512 = \frac{3}{2}\sqrt{3}x^2 + 6x \left( \dfrac{1024}{3\sqrt{3} \, x^2} \right)$
$512 = \dfrac{3\sqrt{3}}{2}x^2 + \dfrac{2048}{\sqrt{3} \, x}$
$\dfrac{3\sqrt{3}}{2}x^3 - 512x + \dfrac{2048}{\sqrt{3}} = 0$
$x = -15.075 ~ \text{(absurd)}, ~ 12.697, ~ \text{and} ~ 2.378$
Try x = 12.697"
$V = \left[ 6 \times \frac{1}{2}(12.697^2)\sin 60^\circ \right] (1.222)$
$V = 511.83 \cong 512 ~ \text{okay}$
Try x = 2.378"
$V = \left[ 6 \times \frac{1}{2}(2.378^2)\sin 60^\circ \right] (34.849)$
$V = 511.995 \cong 512 ~ \text{okay}$
Answer: h = 12.697" or h = 2.378"
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