MATH 1830

Unit 3 Applications of Derivatives

3.8 Optimization in Packaging: Boxes and Other Right Rectangular Prisms

Pre-Class:

3.7 HW

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3.8 Quiz

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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 rectangular package to be sent by a postal service can have a maximum combined length and girth (distance around the package) of 108 inches. Find the dimensions of the package that contains a maximum volume. Assume that the package has square ends.

    $Girth + Length = 108$

    $4x+l=108$

    $l=108-4x$

    $V=lwh$

    $V=l*x*x$

    $V=(108-4x)(x^2)$

    $V=108x^2-4x^3$

    $V'=216x-12x^2$

    Set $V'=0.$

    $216x-12x^2=0$

    $12x(18-x)=0$

    $12x=0$ and $18-x=0$

    $x=0$ and $x=18$

    Use $x=18$ to solve for $l.$

    $ l=108-4x$

    $ l=108-4*18$

    $ l=36$

    The dimensions of the box with the maximum volume are 18 in x 18 in x 36 in.

  2. Your iron works company has been contracted to build a $500 ft^3$ holding tank for a paper company. The square-based, open-top tank is to be made by welding thin stainless steel plates together along their edges. As the production engineer, your job is to find dimensions for the base and height that will make the tank weigh as little as possible. What dimensions do you tell the shop to use?

    $V=lwh$

    $V=x*x*h$

    $500=x^2h$

    $\frac{500}{x^2}=h$

    $SA=lw+2lh+2wh$

    $SA=x^2+2xh+2xh$

    $SA=x^2+4xh$

    $SA=x^2+4x(\frac{500}{x^2})$

    $SA=x^2+\frac{2000x}{x^2}$

    $SA=x^2+\frac{2000}{x}$

    $SA=x^2+2000x^{-1}$

    $SA'=2x-2000x^{-2}$

    Set $SA'=0.$

    $2x-2000x^{-2}=0$

    $2x-\frac{2000}{x^2}=0$

    $\frac{\displaystyle2x}1=\frac{2000}{x^2}$

    $2x^3=2000$

    $x^3=1000$

    $x=10$

    Use $x=10$ to solve for $h.$

    $ \frac{500}{x^2}=h$

    $ \frac{500}{10^2}=h$

    $ \frac{500}{100}=h$

    $ h=5$

    The dimensions of the box with the minimum surface area are 10 ft x 10 ft x 5 ft.

  3. A manufacturer wants to design an open box having a square base and a surface area of 108 $in^2.$ What dimensions will produces a box with a maximum volume?

    $SA=lw+2lh+2wh$

    $108=x^2+2xh+2xh$

    $108=x^2+4xh$

    $108-x^2=4xh$

    $\frac{108-x^2}{4x}=h$

    $V=lwh$

    $V=x*x*h$

    $V=x^2(\frac{108-x^2}{4x})$

    $V=x(\frac{108-x^2}{4})$

    $V=(\frac{108x-x^3}{4})$

    $V=\frac14(108x-x^3)$

    $V=27x-\frac14x^3$

    $V'=27-\frac34x^2$

    Set $V'=0.$

    $27-\frac34x^2=0$

    $\frac{27}{1}=\frac{3x^2}4$

    $3x^2=108$

    $x^2=36$

    $x=6$ and $x=-6$

    Use $x=6$ to solve for $h.$

    $ h=\frac{108-x^2}{4x}$

    $ h=\frac{108-(6^2)}{4*6}$

    $ h=\frac{72}{24}=3$

    The dimensions of the box with the maximum volume are 6 in x 6 in x 3 in.

3.8 Optimization in Packaging: Boxes and Other Right Rectangular Prisms Homework

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 container company is designing an open-top, square-based, rectangular box that will have a volume of 62.5 $cm^3$. Find the dimensions of the box that will minimize the surface area.

    $V=lwh={{w}^{2}}h$

    $62.5={{w}^{2}}h$

    $h=\frac{62.5}{{{w}^{2}}}$

    $SA={{w}^{2}}+2wh+2wh$

    $SA={{w}^{2}}+4wh$

    $S(w)={{w}^{2}}+4w\left( \frac{62.5}{{{w}^{2}}} \right)$

    $S(w)={{w}^{2}}+\frac{250}{w}$

    $S(w)={{w}^{2}}+250w^{-1}$

    $S'(w)=2w-\frac{250}{{{w}^{2}}}$

    Set $S'(w)=0.$

    $0=2w-\frac{250}{{{w}^{2}}}$

    $\frac{250}{{{w}^{2}}}=\frac{2w}1$

    $2{{w}^{3}}=250$

    ${{w}^{3}}=125$

    $w=5$

    Use $w$ to solve for $h.$

    $ 62.5={{5}^{2}}h$

    $ 62.5=25h$

    $ h=2.5$

    The dimensions of the box with the minimum surface area are 5 cm x 5 cm x 2.5 cm.

  2. Find the maximum volume of a box with a square base that has a surface area of 96 $cm^2$.

    Assume a lid.

    $SA=2lw+2lh+2wh$

    $ SA=2({{x}^{2}}+xh+xh) $

    $ 96=2({{x}^{2}}+2xh) $

    $ 48={{x}^{2}}+2xh $

    $ 48-{{x}^{2}}=2xh $

    $ \frac{48-{{x}^{2}}}{2x}=h $

    $V=l\cdot w\cdot h$

    $ V(x)=x(x)\left( \frac{48-{{x}^{2}}}{2x} \right) $

    $ V(x)=\frac{x(48-{{x}^{2}})}{2} $

    $ V(x)=\frac{48x-{{x}^{3}}}{2}$

    $ V(x)=\frac{48x}{2}-\frac{{{x}^{3}}}{2} $

    $V(x)=24x-0.5{{x}^{3}}$

    $V'(x)=24-1.5{{x}^{2}}$

    $0=24-1.5{{x}^{2}}$

    $1.5{{x}^{2}}=24$

    ${{x}^{2}}=\frac{24}{1.5}$

    ${{x}^{2}}=16$

    $x=4$

    $h=\frac{48-16}{2(4)}=4$

    The dimensions of the box that maximize the volume are 4 cm x 4 cm x 4 cm.

    The maximum volume is 64 $cm^3$

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