iconEuler Examples

Paraboloid with Maximal Volume

by R. Grothmann

We search the paraboloid of maximal volume, which fits inside
a cone.

A paraboloid is a parabola y=x^2 rotated around the y-axis. Here
is an image.
>plot3d("1-x^2",xmin=0,xmax=1,rotate=2); insimg;

Paraboloid wth maximal volume

We want to fit such a paraboloid under the cone, which is a the
rotated function 1-|x|.
>plot3d("1-x",xmin=0,xmax=1,rotate=2); insimg;

Paraboloid wth maximal volume

We first assume, that the parabola has the form ax^2+b.
>:: function p(x,a,b) := a*x^2+b
                                             2
                            p(x, a, b) := a x  + b

It should touch 1-|x|. So we find all parabolas that touch this
function.
>:: solve(diff(p(x,a,b),x)=-1,x), sb := solve(at(p(x,a,b)=1-x,%),a)
                                          1
                                  [x = - ---]
                                         2 a


                                         1
                                 [a = -------]
                                      4 b - 4

We get a family of parabolas.
>:: pt(x,b) := at(p(x,a,b),sb)
                                           2
                                          x
                            pt(x, b) := ------- + b
                                        4 b - 4

Let us check with plots.
>b=(0.1:0.1:0.9)'; plot2d("1-abs(x)",-1,1);  ...
>  plot2d(mxm("pt(x,b)"),-1,1,add=1); insimg;

Paraboloid wth maximal volume

Now we need to compute the volume of the rotated parabola for
y>0. We need to determine the radius of the paraboloid for specific
y.
>:: yinv := solve(pt(x,b)=y,x)
                                 2                             2
        [x = - 2 sqrt(b y - y - b  + b), x = 2 sqrt(b y - y - b  + b)]

Then we find the volume.
>:: define(F(b),integrate(%pi*rhs(yinv[1])^2,y,0,b))
                                             3    2
                           F(b) := - 2 %pi (b  - b )

And maximize it.
>:: solve(diff(F(b),b)=0,b), F(b) | %[1], float(%)
                                     2
                                [b = -, b = 0]
                                     3


                                     8 %pi
                                     -----
                                      27


                               0.93084226773031

Plot this specific parabola.
>plot2d("1-abs(x)",-1,1); plot2d(mxm("pt(x,2/3)"),add=1); insimg;

Paraboloid wth maximal volume

This is our solution.
>:: pt(x,2/3)
                                          2
                                   2   3 x
                                   - - ----
                                   3    4

Now we solve a similar problem. We request, that the parabola must
fit under the cone function 1-|x|, and then turn such a parabola
around the y-axis.

It is not obvious that the maximal volume is achieved, if the
parabola is symmetric. However, think of a parabola which touches
1-|x| in some point x>0. We can assume that the vertex of the parbola
is on the positive side. Then we can show that there is a larger
parabola, which touches 1-|x| in the same point x, and is symmetric
with respect to the y-axis.

However, for the following, we solve the general case numerically,
and assume that the parabola has two zeros a, b with

-1 < -b < a < b < 1

and touches 1-x somewhere in [0,b].

>:: function p(x) := c*(x-a)*(x-b)
                           p(x) := c (x - a) (x - b)

>:: solve(diff(p(x),x)=-1,x); sc := rhs(solve(at(p(x)=1-x,%),c)[1])
                      2 sqrt((a - 1) b - a + 1) - b - a + 2
                    - -------------------------------------
                                  2            2
                                 b  - 2 a b + a

>:: define(ps(x),at(p(x),c=sc))
                  (2 sqrt((a - 1) b - a + 1) - b - a + 2) (x - a) (x - b)
       ps(x) := - -------------------------------------------------------
                                       2            2
                                      b  - 2 a b + a

The functions pt are the set of such parabolas. Let us plot a
few samples.
>plot2d("1-abs(x)",r=1); ...
>a=(-0.9:0.1:0)'; b=0.9; plot2d(mxm("ps(x)"),add=1); insimg;

Paraboloid wth maximal volume

Let us plot such a rotated parabola.
>a=0.8; b=-0.3; plot3d("::ps(x)",xmin=0,xmax=0.8,rotate=2); insimg;

Paraboloid wth maximal volume

To determine the area of the rotated parabola, we integrate with
polar coordinates.
>:: integrate(2*%pi*x*ps(x),x); V(a,b) := factor(at(%,x=b))
                         3
                    %pi b  (b - 2 a) (b - 2 sqrt((a - 1) (b - 1)) + a - 2)
       V(a, b) := - ------------------------------------------------------
                                                   2
                                          6 (b - a)

The maximum is the same as before. We take a=-y*x, b=x with 0<x<1 and
0<y<1. The plot shows a maximum with y=1. So we have two symmetric
zeros, and get the same case as above.
>plot3d(mxm("V(x,-y*x)"), ...
>  xmin=0.1,xmax=1,ymin=0,ymax=1,user=1,xlabel="a",ylabel="b",scale=1); ...
>insimg;

Paraboloid wth maximal volume

It is not easy to show that V(a,b) increases monotonically as a
decreases from 0 to -b. But if we assume this is true, we get the
same maximum as before.

However, we have to use Euler's numerical algorihms, since Maxima
cannot solve this instantly.
>xmax=fmax(&V(-x,x),0,1); &V(-x,x)(xmax)
0.9308422677303
>

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