the part of the parabola y=4-x^2 from x=0 to x=2 is rotated about the y axis to form a surface. A cone is inscribed in the resulting paraboloid with its vertex at the origin and its base touching the parabola. At what radius and altitude does the max volume occur? What is the max volume?

Question

the part of the parabola y=4-x^2 from x=0 to x=2 is rotated about the y axis to form a surface. A cone is inscribed in the resulting paraboloid with its vertex at the origin and its base touching the parabola.  At what radius and altitude does the max volume occur?  What is the max volume?

anonymous · Answer

What is your question?

anonymous · Answer

the part of the parabola y=4-x^2 from x=0 to x=2 is rotated about the y axis to form a surface. A cone is inscribed in the resulting paraboloid with its vertex at the origin and its base touching the parabola.  At what radius and altitude does the max volume occur?  What is the max volume?

anonymous · Answer

Yes that makes sense. Thank you so much for your help!

anonymous · Answer

Wait, There is a small correction.

anonymous · Answer

OK. Thanks.

anonymous · Answer

It does not change the answer. I will rewrite the new version.

anonymous · Answer

$$
V(x) =\frac 1 3 x(4-x^2)=\frac{4 x}{3}-\frac{x^3}{3}\
V'(x)=\frac{4}{3}-x^2\
V''(x)=-3 x\
V'(x) = 0, \implies x=\pm \frac 2 {\sqrt 3}
$$

anonymous · Answer

We take the positive value of x and at that point the second derivative is negative so we have a maximum

anonymous · Answer

Thank you!

anonymous · Answer

YW