Here is a question that I have, but I have no idea where to do go from here. Here is the question:
The vase company designs a new vase that is shaped like a cylinder on the bottom with a cone on top. The catalog states that the width is $12$ cm and the total height is $42$ cm. What would the height of the cylinder part have to be in order for the total volume to be $1224 \pi$ $\mathrm{cm}^3$.
The three equations that I got to help me solve this problem are:
Volume of a cylinder: $V=\pi r^2h$
Volume of a cone: $V=\frac{\pi r^2h}{3}$
Volume of a sphere: $V=\frac{4\pi r^3}{3}$
$h$ =height
$r$=radius
How can I start this problem off and solve it?
$\endgroup$ 03 Answers
$\begingroup$The diameter of the cylinder is $12$, so the radius is $6$. Let $a$ be the height of the cylinder, and $b$ the height of the cone. We are told that $$a+b=42.$$
The volume of the cylinder is, by one of the formulas you quoted, equal to $36\pi a$, and the volume of the cone, by another of the formulas, is $12\pi b$. Since the combined volume is $1224\pi$, we obtain $$36a+12b=1224.$$ We now have two linear equations in the two unknowns $a$ and $b$. Solve.
$\endgroup$ $\begingroup$Let the height of the cylinder be $h$ cm.
Then the height of the cone becomes $(42-h)$ cm.
The diameter of both the cone and the cylinder are given to be $12$ cm.
The total volume should be the sum of the cylinder volume and the cone volume, i.e., if $V_T$ is the total volume and $V_\text{cylinder}$ is the cylinder volume and $V_\text{cone}$ is the cone volume, then
$V_\text{T}=V_\text{cylinder}+V_\text{cone}$
$1224\pi$ = $\pi r^2h+\frac{\pi r^2(42-h)}{3}$
Dividing the above equation by $\pi$ and setting $r=6$ cm, we have
$1224$ = $(6)^2h+\frac{(6)^2(42-h)}{3}$
The above yields
$h=30$ cm
$\endgroup$ 4 $\begingroup$Call the height of the cylinder $h_1$ and that of the cone $h_2$. The height of the vase is given:
$$h_1+h_2=H = 42 \text{cm}$$
The volume of the vase is also given:
$$\pi R^2 h_1 + \frac13 \pi R^2 h_2 = V = 1224 \pi \text{cm}^3$$
where $R = 6 \text{cm}$. You now have 2 equations and 2 unknowns; solve.
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