Vase Maker Makes a Vase

Contour of vase is a cubic polynomial we can write that as:

$P(x)=ax^3+bx^2+cx+d$

We know that $P(x)$ passes through points $(1,2)$ and $(3,1)$ , we can write this as:

$P(1)=2$

$P(3)=1$

We also know that those points represent widest and narrowest point, meaning that those points are local maximum/minimum, we can write it as:

$P'(1)=0$

$P'(3)=0$

where:

$P'(x)=\frac{dP(x)}{dx}=3ax^2+2bx+c$

By putting the values we get following system of linear equations

$\begin{vmatrix} a+b+c+d=1 \\ 27a+9b+3c+d=1 \\ 3a+2b+c=0\\ 27a+6b+c=0 \end{vmatrix}$

By solving it we get:

$\begin{vmatrix} a=\frac{1}{4} \\ b=-\frac{3}{2} \\ c=\frac{9}{4}\\ d=1 \end{vmatrix}$

Now when we have found our polynomial to be $P(x)=\frac{1}{4}x^3-\frac{3}{2}x^2+\frac{9}{4}x+1$ . We can find volume of the vase as:

$\displaystyle V=\int_{0}^{4}\pi P(x)^2 dx$

This is an easy polynomial integral and by evaluating it we get:

$V=\frac{332\pi}{35}\approx29.8 \text{ cubic feet} .\square$

Slight variation on Miloje Đukanović's approach.

The contour of the vase must have the points $(1,2)$ and $(3,1)$ as extrema. Since the contour is a cubic, this means that its derivative must have roots at $x = 1, 3$ . That is, $P^\prime(x) = a(x - 1)(x - 3).$ Integrating we have, $P(x) = a\left(\frac{1}{3}x^3 - 2x^2 + 3x\right) + c,$ where $a,c$ are real constants. The turning points of this cubic are at the required x-coordinates, but we still need to ensure that they are at the required y-coordinates. Evaluating at $x = 3$ gives $P(3) = a(9 - 18 + 9) + c = c = 1.$ Evaluating at $x = 1$ gives $P(1) = a\left(\frac{1}{3} - 2 + 3\right) + 1 = \frac{4}{3}a + 1 = 2,$ which solves to $a = \frac{3}{4}$ . Therefore, the contour function for the vase is $P(x) = \frac{1}{4}x^3 - \frac{3}{2}x^2 + \frac{9}{4}x + 1$ This can then be integrated to find the required volume.