# what's the greatest volume of a cylinder using calculus?

I have a rectangle that has the perimeter of 38cm. I need to make this rectangle into a baseless cylinder and find the greatest volume of it, by deriving.

so far I came with this: for the rectangle the length is 19 - width

L=(19-w)

W= w

then subbed it into finding the radius of the cylinder:

$r= circumference/2\pi$

$r= w/2\pi$

and finally subbed everything into the volume equation:

$V=\pi r^2 h$

$v= \pi (19-(w/2\pi)) (19 -w)$

but now I'm having trouble to derive it and find the greatest volume.

• Have you attempted the problem yourself? You should include any work you have done on the problem and indicate where you are stuck so that you receive responses that address the specific difficulties you are encountering. – N. F. Taussig Sep 1 '15 at 8:50

The width will be the circumference of the base circle, hence its radius will be $\dfrac{w}{2\pi}$ and its area $\dfrac{w^2}{4\pi}$, and the volume of the cylinder: $$V=\smash{\dfrac{w^2}{4\pi }} (19-w)$$ Now$$\frac{\mathrm d\mkern0.5mu V}{\mathrm d\mkern0.5mu w}=\smash{\dfrac1{4\pi}}\bigl(2w(19-w)-w^2\bigr)=\smash{\dfrac{w(38-3w)}{4\pi}}$$ Furthermore, the signs of the derivative on $\mathbf R$ are $$\begin{matrix} w&&0&&\frac{38}3&\\ \hline \frac{\mathrm d\mkern0.5mu V}{\mathrm d\mkern0.5mu w}&-&0&+&0&- \end{matrix}$$ hence there is a local maximum at $\dfrac{38}3$ and this maximum is equal to $$V_\text{max}=\frac1{4\pi}\frac{19^3}{3^3}=\frac{6859}{108\pi}.$$
• Wouldn't the width $w$ be the circumference of the base circle, so that its area will be $\pi\left(\frac{w}{2\pi}\right)^2=\frac{w^2}{4\pi}$? – Alijah Ahmed Sep 1 '15 at 9:56
• Your welcome - the edit needed was just to move $\pi$ to the denominator. (+1) – Alijah Ahmed Sep 1 '15 at 10:17