2.7 Optimization in Packaging: Cans and Other Right Cylinders
2.7 Video
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.
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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. What is the minimum surface area?
$ 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 $SA'=0.$
$4\pi r-\frac{710}{{{r}^{2}}}=0$
$4\pi r=\frac{710}{{{r}^{2}}}$
$\frac{4\pi r}{1}=\frac{710}{{{r}^{2}}}$
$4\pi {{r}^{3}}=710$
${{r}^{3}}=\frac{710}{4\pi }$
${{r}^{3}}=56.5$
$r=3.84$
$h=\frac{355}{\pi {{\left( 3.84 \right)}^{2}}}=7.66$
$ SA=2\pi {{r}^{2}}+710{{r}^{-1}}$
$ SA=2\pi {{(3.84)}^{2}}+710{{(3.84)}^{-1}}=277.55$
The surface area of the can will be minimized when the radius is 3.84 cm and the height is 7.66 cm, and the minimum surface area will be 277.55 square cm.
2.7 Lecture
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You have 320 $cm^2$ of aluminum to make a cylinder shaped coke can. Find the dimensions that would produce a maximum volume. What is the 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 $V'=0.$
$160-3πr^2=0$
$160=3πr^2$
$r^2=\frac{160}{3π}$
$r=\sqrt{\frac{160}{3π}}$
$r=\pm4.12$
$h= \frac{320-2πr^2}{2πr}$
$h= \frac{320-2π(4.12)^2}{2π(4.12)}$
$h= 8.24$
$V=160r-πr^3$
$V=160(4.12)-π(4.12)^3=439.49$
The volume of the can will be maximized when the radius is 4.12 cm and the height is 8.24 cm, and the maximum volume is 439.49 cubic cm.
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What are the dimensions of the lightest open-top right cylindrical container that can hold a volume of 1000 $cm^3$? What is the minimum surface area?
$ 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 $SA'=0.$
$2\pi r-\frac{2000}{{{r}^{2}}}=0$
$2\pi r=\frac{2000}{{{r}^{2}}}$
$\frac{2\pi r}{1}=\frac{2000}{{{r}^{2}}}$
$2\pi {{r}^{3}}=2000$
${{r}^{3}}=\frac{2000}{2\pi }$
${{r}^{3}}=\frac{1000}{\pi }$
$\sqrt[3]{\frac{1000}{\pi}}$
$r=6.83$
$h=\frac{1000}{\pi (6.83)^{2}} =6.82$
$ SA=\pi {{r}^{2}}+2000{{r}^{-1}}$
$ SA=\pi {{(6.83)}^{2}}+2000{{(6.83)}^{-1}}=439.38$
The surface area of the can will be minimized when the radius is 6.83 cm and the height is 6.82 cm, and the mimimum surface area is 439.38 square cm.
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A cylinderical container is to be made with 3000 $cm^2$ of sheet metal. What dimensions would result in the maximum volume? What is the maximum volume? Assume a lid.
$SA=2πr^2+2πrh$
$3000=2πr^2+2πrh$
$ 3000-2πr^2=2πrh$
$ \frac{3000-2\mathrm{πr}^2}{2πr}=h$
$ V=πr^2h$
$ V=πr^2(\frac{3000-2\mathrm{πr}^2}{2πr})$
$ V=r(\frac{3000-2\mathrm{πr}^2}{2})$
$ V=1500r-πr^3$
$ V'=1500-3πr^2$
$ 0=1500-3πr^2$
$ 1500=3πr^2$
$ 500=πr^2$
$ \frac{500}π=r^2$
$ \sqrt{\frac{500}{π}}=r$
$ r=\pm12.62$
$ h=\frac{3000-2π(12.62)^2}{2π(12.62)}=25.21$
$ V=1500r-πr^3$
$ V=1500(12.62)-π(12.62)^3$
The maximum volume of 12,615.66 cubic cm occurs when the radius is 12.62 cm and the height is 25.21 cm.
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A cylindrical can has a volume of 900 cm³. The metal costs $15.50/cm². What dimensions produce a can with minimized cost? What is the cost of making the can?
$V=\pi {{r}^{2}}h$
$ 900=\pi {{r}^{2}}h$
$ h=\frac{900}{\pi {{r}^{2}}}$
$ SA=2\pi {{r}^{2}}+2\pi rh$
$ S(r)=2\pi {{r}^{2}}+2\pi r\left( \frac{900}{\pi {{r}^{2}}} \right)$
$ S(r)=2\pi {{r}^{2}}+\frac{1800}{r}$
$ S(r)=2\pi {{r}^{2}}+1800{{r}^{-1}}$
$ S'(r)=4\pi r-\frac{1800}{{{r}^{2}}}$
$ 0=4\pi r-\frac{1800}{{{r}^{2}}}$
$ \frac{1800}{{{r}^{2}}}=4\pi r$
$ 4\pi {{r}^{3}}=1800$
$ {{r}^{3}}=\frac{1800}{4\pi }$
$ {{r}^{3}} \approx 143.239448$
$ r\approx 5.23\text{ cm}$
$ h=\frac{900}{\pi {{(5.23)}^{2}}}=10.47\text{ cm}$
Minimum Surface Area
$ SA=2\pi {{r}^{2}}+2\pi rh$
$ SA=2\pi {{\left( 5.23 \right)}^{2}}+2\pi \left( 5.23 \right)\left( 10.47 \right)$
$ SA=515.92\text{ }c{{m}^{2}}$
Cost of the can
$ =\$15.50(SA)$
$=\$15.50(515.92)$
$=7996.76$
The minimum surface area occurs when the radius is 5.23 cm and the height is 10.47 cm.
The cost of making the can is \$7996.76.
2.7 Group Work
2.7 Additional Practice
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.