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 May 3rd, 2011, 09:13 AM #1 Newbie   Joined: May 2011 Posts: 9 Thanks: 0 Largest and smallest volumes Hi I need some help in a differentiation question but I dont need much help in the differentiating part but only with starting the question off. The surface area of a cylinder is given which is 60cm. However the radius and height is not given. a) would we need to find the h and r to find the largest possible volume? And if you do how do you find it?? b)I know to find the maximum volume you differentiate and then put the value back into the original equation for surface area to find the maximum volume. but please correct me if i am wrong. Any help would be appreciated.
 May 3rd, 2011, 09:42 AM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Largest and smallest volumes You are given $2\pi r(h+r)=60$. Rearrange that to find h in terms of r, and substitute the result into the formula for volume $\pi r^2h$ to get a function of r alone, V(r). Then differentiate and solve V'(r)=0.
 May 3rd, 2011, 09:50 AM #3 Newbie   Joined: May 2011 Posts: 9 Thanks: 0 Re: Largest and smallest volumes Can the solutions be found by rearranging or can you make it into a quadratic equation?
 May 3rd, 2011, 10:14 AM #4 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Largest and smallest volumes You turn the two-variable function V(h,r) into a single-variable function V(r), and then V'(r)=0 is a quadratic equation (and an easy one). Can you find h in terms of r?
 May 3rd, 2011, 10:27 AM #5 Newbie   Joined: May 2011 Posts: 9 Thanks: 0 Re: Largest and smallest volumes Hi Yes Ive found h in terms of r which is 30-pi r^2/pi r substituting h into the volume of cylinder formula gives me -pi r^3 + 30 Is that correct?
 May 3rd, 2011, 10:38 AM #6 Newbie   Joined: May 2011 Posts: 9 Thanks: 0 Re: Largest and smallest volumes when this is differentiated r=0 and when this is put into the original equation it doesnt make sense!
 May 3rd, 2011, 12:02 PM #7 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Largest and smallest volumes You made a small error in finding V(r): $h\,=\,\frac{30}{\pi r}\,-\,r\\V\,=\,\pi r^2h\,=\,30r-\pi r^3$
 May 3rd, 2011, 03:50 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs Re: Largest and smallest volumes The formula for the total surface area A of a closed right-circular cylinder having radius r and height h is: $A=2\pi r^2+2\pi rh$ Solving for h, we have: $h=\frac{A-2\pi r^2}{2\pi r}$ The formula for the volume V of the same cylinder is: $V=\pi r^2h$ Substituting for h, we have $V=\pi r^2\left(\frac{A-2\pi r^2}{2\pi r}\right)=r\left(\frac{A-2\pi r^2}{2}\right)=\frac{Ar}{2}-\pi r^3$ To find the maximum volume, we differentiate V with respect to r and equate to zero: $\frac{dV}{dr}=\frac{A}{2}-3\pi r^2=0$ $r=\sqrt{\frac{A}{6\pi}}$ To verify this value of r gives a maximum for V, we can use the second derivative test: $\frac{d^2V}{dr^2}=-6\pi r$ We see V''(r) will be negative, meaning concavity is downward, demonstrating the critical value for r gives a maximum for V. Now we can find the critical value of h in terms of r: $h=\frac{A-2\pi$$\sqrt{\frac{A}{6\pi}}$$^2}{2\pi$$\sqrt{\frac {A}{6\pi}}$$}=\frac{A-\frac{A}{3}}{2\pi\sqrt{\frac{A}{6\pi}}}=\frac{\fra c{2A}{3}}{2\pi\sqrt{\frac{A}{6\pi}}}=$ $\text{ }\frac{A}{3\pi\sqrt{\frac{A}{6\pi}}}=\frac{A}{\sqr t{\frac{3A\pi}{2}}}=\sqrt{\frac{2A}{3\pi}}=2\sqrt{ \frac{A}{6\pi}}=2r$ Thus, we see that a cylinder having the maximum volume for a given surface area, or conversely, the minimum surface area for a given volume will have its diameter equal to its height. Placing the critical value for r into V to find the maximum volume, we find $V=\frac{Ar}{2}-\pi r^3=\sqrt{\frac{A}{6\pi}}$$\frac{A}{2}-\pi \(\sqrt{\frac{A}{6\pi}}$$^2\)=\sqrt{\frac{A}{6\pi} }$$\frac{A}{3}$$=\frac{A^{\frac{3}{2}}}{3\sqrt{6\p i}}$ Now we have r, h, and V in terms of the surface area A.

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