Would you use a proportion to solve this problem? A math class is working with similar figures. Mr. Kole told the class that the surface area of the cones are 180 and 320 square units, and the volume of the smaller cone is 151 cubic units. He challenged his class to find the volume of the bigger cone. What is the volume of the bigger cone?
I tried to use the proportion 320/180 = x/151 to solve the problem. But my answer did not match the correct answer, which is 358 cubic units.
How would you solve this problem?
 A: Since the cones are similar the ratio of their areas is equal to the square of corresponding lines.  (For example, the height or the radius of the base.)  The ration of those lines is $$a=\sqrt{{320\over180}}=\frac43$$  The ratio of the volume is $a^3$ so the volume of the large cone is $$\left(\frac43\right)^3\cdot151$$
A: Well you could use the formulas for the are and volume of cones to figure everything out and I guess that will be enough to solve the problem. However the problem is much more simple than that. 
The area scales as a square of the radius $A\propto r^2$ and the volume as the cube $V\propto r^3$. In particular $A^3\propto r^6,\, V^2\propto r^6$, then the quantity 
$$
\frac{A^3}{V^2}
$$
is an invariant for similar cones (or any other figure for that mater). And we have
$$
\frac{320^3}{x^2} = \frac{180^3}{151^2}
$$
This should be alright, however I am getting $x=358$ which doesn't seem to be the answer you are getting. Could you please check the numbers?
A: No, surface area is not directly proportional to volume.
I would try to construct a cone with surface area $180$ and volume $151$ using the formulas for surface area and volume, in particular keeping track of the ratio between the cone's height and radius.
Then I would construct a cone with surface area $320$ and the same ratio between its height and radius. Finally, apply the volume formula to this larger cone.
A: Volumes are proportional to cube and areas to square of linear dimension in similitude. 
$$ V = k_1 L^3,\, A = k_2 L^2,\, V = k_3 A^{\frac32} $$
Accordingly the larger Volume is
$$ V_2 = V_1 \big(\frac{320}{180}\big)^ {1.5} =151\times 64/ 27 \approx 357.926 $$
A: The similarity of two cones imply:
$$\frac{R_1}{R_2}=\frac{l_1}{l_2}=\frac{h_1}{h_2} \Rightarrow \frac{R_1}{R_2}=\frac{R_1+l_1}{R_2+l_2}$$
The areas of cones:
$$S_1=\pi R_1l_1+\pi R_1^2=180;\\
S_2=\pi R_2l_2+\pi R_2^2=320;\\
\frac{S_1}{S_2}=\frac{\pi R_1(l_1+R_1)}{\pi R_2(l_2+R_2)}=\left(\frac{R_1}{R_2}\right)^2=\frac{180}{320} \Rightarrow \left(\frac{R_1}{R_2}\right)^3=\left(\frac{180}{320}\right)^{3/2}$$
The volumes of cones:
$$V_1=\pi R_1^2h_1=151;\\
V_2=\pi R_2^2h_2;\\
\frac{V_1}{V_2}=\frac{\pi R_1^2h_1}{\pi R_2^2h_2}=\left(\frac{R_1}{R_2}\right)^3=\left(\frac{180}{320}\right)^{3/2}=\frac{151}{V_2} \Rightarrow V_2\approx 357.93.$$
