C1 biscuit tin

Confused on the last part. Surely the smallest sphere which the tin fits is one which radius tends to 0?? Not sure how to approach this part, or how to think about it.

Confused on the last part. Surely the smallest sphere which the tin fits is one which radius tends to 0?? Not sure how to approach this part, or how to think about it.

The tin is supposed to be inside the sphere.
You could find the longest 'radius' of a cylinder, draw a good sketch and do it that way, but that's not the method they want.

Can't think of another method right now :s

Ah, I see. Let me know if this is right:

The cylinder has radius $r$ and height $h$. From previous parts I worked out $h=2r$.

The 'radius' from the centre of the cylinder, which is also the centre of the sphere, would be $R=\sqrt{r^2+(\frac{1}{2}h)^2}$ and R is the smallest radius of the sphere. Using substitution for h, I get $R=r\sqrt2$.

So the volume of the sphere would be $\displaystyle V_s=\frac{4}{3} \pi R^3=\frac{2^2}{3} \pi (r\sqrt2)^3=\frac{2^{7/2}}{3}\pi r^3$

Ah, I see. Let me know if this is right:

The cylinder has radius $r$ and height $h$. From previous parts I worked out $h=2r$.

The 'radius' from the centre of the cylinder, which is also the centre of the sphere, would be $R=\sqrt{r^2+(\frac{1}{2}h)^2}$ and R is the smallest radius of the sphere. Using substitution for h, I get $R=r\sqrt2$.

So the volume of the sphere would be $\displaystyle V_s=\frac{4}{3} \pi R^3=\frac{2^2}{3} \pi (r\sqrt2)^3=\frac{2^{7/2}}{3}\pi r^3$

Might want to put that in terms of V, which is one of the given quantities.