MATH 1830

3.7 Optimization in Packaging: Cans and Other Right Cylinders

Pre-Class:

• Complete 3.6 Homework assignment: check and correct.
• Take notes on the videos and readings (use the space below).
• Complete the 3.7 Pre-Class Quiz.

Write a mathematical function for each of the following and then find the requested information.

Be sure to include a properly labeled diagram (if applicable) and variable statements. State the restrictions on the independent variable.

1. A cylindrical soup can has a volume of 355 $cm^3.$ Find the dimensions (radius and height) that minimize the surface area of the can.

   $ V=355$ $c{{m}^{3}} $

   $ V=355=\pi {{r}^{2}}h $

   $ h=\frac{355}{\pi {{r}^{2}}} $

   $ SA=2\pi {{r}^{2}}+2\pi rh$

   $ SA=2\pi {{r}^{2}}+2\pi r\left( \frac{355}{\pi {{r}^{2}}} \right)$

   $ SA=2\pi {{r}^{2}}+\frac{710}{r}$

   $ SA=2\pi {{r}^{2}}+710{{r}^{-1}}$

   $SA'=4\pi r-710r^{-2}$

   $SA'=4\pi r-\frac{710}{{{r}^{2}}}$

   Set $SA'=0.$

   $4\pi r-\frac{710}{{{r}^{2}}}=0$

   $4\pi r=\frac{710}{{{r}^{2}}}$

   $\frac{4\pi r}{1}=\frac{710}{{{r}^{2}}}$

   $4\pi {{r}^{3}}=710$

   ${{r}^{3}}=\frac{710}{4\pi }$

   ${{r}^{3}}=56.5$

   $r=3.84$

   $h=\frac{355}{\pi {{\left( 3.84 \right)}^{2}}}=7.66$

   The surface area of the can will be minimized when the radius is 3.84 cm and the height is 7.66 cm.

2. You have 320 $cm^2$ of aluminum to make a cylinder shaped coke can. Find the dimensions that would produce a maximum volume.

   $ SA=2πr^2+2πrh$

   $ 320=2πr^2+2πrh$

   $ 320-2πr^2=2πrh$

   $ \frac{320-2πr^2}{2πr}=h$

   $ V=πr^2h$

   $ V=πr^2(\frac{320-2πr^2}{2πr})$

   $ V=r(\frac{320-2πr^2}{2})$

   $V=160r-πr^3$

   $V'=160-3πr^2$

   Set $V'=0.$

   $160-3πr^2=0$

   $160=3πr^2$

   $r^2=\frac{160}{3π}$

   $r=\sqrt{\frac{160}{3π}}$

   $r=\pm4.12$

   $h= \frac{320-2πr}{2πr}$

   $h= \frac{320-2π(4.12)}{2π(4.12)}$

   $h= 11.36$

   The volume of the can will be maximized when the radius is 4.12 cm and the height is 11.36 cm.

3. What are the dimensions of the lightest open-top right cylindrical container that can hold a volume of 1000 $cm^3$?

   $ V=1000$ $c{{m}^{3}} $

   $ V=1000=\pi {{r}^{2}}h $

   $ h=\frac{1000}{\pi {{r}^{2}}} $

   $ SA=\pi {{r}^{2}}+2\pi rh$

   $ SA=\pi {{r}^{2}}+2\pi r(\frac{1000}{\pi {{r}^{2}}} )$

   $ SA=\pi {{r}^{2}}+\frac{2000}{r}$

   $ SA=\pi {{r}^{2}}+2000{{r}^{-1}}$

   $SA'=2\pi r-{2000}{{{r}^{-2}}}$

   $SA'=2\pi r-\frac{2000}{{{r}^{2}}}$

   Set $SA'=0.$

   $2\pi r-\frac{2000}{{{r}^{2}}}=0$

   $2\pi r=\frac{2000}{{{r}^{2}}}$

   $\frac{2\pi r}{1}=\frac{2000}{{{r}^{2}}}$

   $2\pi {{r}^{3}}=2000$

   ${{r}^{3}}=\frac{2000}{2\pi }$

   ${{r}^{3}}=\frac{1000}{\pi }$

   $\sqrt[3]{\frac{1000}{\pi}}$

   $r=6.83$

   $h=\frac{1000}{\pi (6.83)^{2}} =6.82$

   The surface area of the can will be minimized when the radius is 6.83 cm and the height is 6.82 cm.