Can we maximise the volume of this metal tin?

Can we maximise the volume of this metal tin?

Question

A cylindrical tin and its tight-fitting cylindrical lid (overlapping the tin by $2\,\text{cm}$) are to be made from thin sheet metal. When made, the total area of sheet metal used is $350\pi\,\text{cm}^{2}$. The base radius of the tin is $x\,\text{cm}$ and its height is $y\,\text{cm}$ as shown in the diagram. Show that, neglecting the thickness of the metal, $x^2+xy+2x=175.$

Deduce that the volume, $V\,\text{cm}^{3}$, of the tin is given by $V= \pi(175x-2x^2-x^3).$

If $x$ may vary, find the values of $x$ and $y$ for which $V$ has its maximum value.

Suggestion

Can you work out the surface area and volume of this tin and lid in terms of $x$ and $y$? Now can you eliminate $y$?

Solution

We know that the surface area of the tin is

$350\pi\,\text{cm}^{2}$

. In terms of

$x$

and

$y$

, the surface area of the tin is:

$$\begin{align*} 350\pi &=\text{area of base} +\text{area of side}+\text{area of side of lid}+\text{area of top of lid} \\ &= \pi x^2+2\pi x y+2\pi x \times2+\pi x^2 \\ &= 2\pi x^2+2\pi xy +4\pi x, \end{align*}$$

and dividing through by

$2\pi$

we get

$$\begin{equation} 175=x^2+xy+2x. \end{equation}$$

The volume of the tin is $V= \pi x^2 y.$

We can eliminate $y$ by rearranging equation $(1)$ to find that $xy=175-x^2-2x$, and so $V=\pi x^2 y=\pi x(xy)=\pi x(175-x^2-2x)=\pi(175x-x^3-2x^2).$

We differentiate $V$ with respect to $x$, giving $\frac{dV}{dx}=\pi(175-3x^2-4x)=0,$ so to find stationary points of $V$, we must solve $3x^2+4x-175=0.$

We can factorise this equation as $(3x+25)(x-7)=0$ (or use the quadratic formula).

Since $x$ is the radius of the tin, it must be positive, so $x=7$.

We have that $\frac{d^2V}{dx^2}=\pi(-6x-4),$ which is negative when $x$ is positive, so $x=7$ gives a local maximum for $V$.

Furthermore, as there are no other stationary points for $V$ with $x>0$, this must give the maximum possible value for $V$ for $x\ge0$.

We need to find the corresponding $y.$ We saw in the previous part that $xy=175-x^2-2x,$ so that $7y=175-49-14=112,$ and hence $y=16$.