The volumes of two cylindrical cans of the same shape vary directly as the cubes of their radii. If a can with a six-inch radius holds pints, how many gallons will a similar can with a 24-inch radius hold?

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Answer 1 · 2018-07-30T02:33:59

$V_{2} = 8 gallons$ $color(brown)(V_2 = 8 " gallons"$

$Volume of cylinder = V = π r^{2} h$ $"Volume of cylinder " = V = pi r^2 h$

Let $r_{1}, h_{1}$ $r_1, h_1$ be the radius and height of the first cylinder and $r_{2}, h_{2}$ $r_2, h_2$ the second.

$r_{1} = 6 inch, r_{2} = 24 inch, V_{1} = 1 pint$ $color(purple)(r_1 = 6 " inch", r_2 = 24 " inch", V_1 = 1 " pint"$

$V_{1} = π r_{1}^{2} h_{1}, V_{2} = π r_{2}^{2} h_{2}$ $V_1 = pi r_1^2 h_1, V_2 = pi r_2 ^2 h_2$

$\frac{V_{2}}{V_{1}} = \frac{r_{2}^{3}}{r_{1}^{3}} = \frac{24^{3}}{6^{3}}, given$ $V_2 / V_1 = r_2 ^3 / r_1 ^3 = 24^3 / 6^3, " given"$

$1 g a l l o n = 8 π n t s$ $1 gallon = 8 pints$

$:. V_2 = (24/6)^3 * V_1 = (24/6)^3 * 1 = 4^3 = 64 " pints"$

$color(brown)(V_2 = 64 / 8 = 8 " gallons, as 8 pints = 1 gallon"$