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Applied maximum and minimum problems

  1. Oct 27, 2010 #1
    Q: A box will be built with a square base and an open top. Material for the base costs $8 per square foot, while material for the sides costs $2 per square foot. Find the dimensions of the box of maximum volume that can be built for $2400.

    A:
    $8[tex]=c1/b[/tex]
    $2[tex]=c2/s[/tex]
    [tex]C=c1+4c2=8b+(4)(2)s=8b+8s=8x^{2}+8xy=8(x^{2}+xy)=2400[/tex]
    [tex]V=x^{2}y[/tex]

    [tex]D_{x}C=?:[/tex]
    [tex]D_{x}C=8x[2x+y+xD_{x}y]=0[/tex]

    [tex]D_{x}y=?:[/tex]
    [tex]2x+y+xD_{x}y=0[/tex]
    [tex]xD_{x}y=-2x-y[/tex]
    [tex]D_{x}y=-2-y/x[/tex]

    [tex]D_{x}V=?:[/tex]
    [tex]D_{x}V=2xy+x^{x}D_{x}y=2xy-(2+y/x)x^{2}[/tex]
    [tex]=2xy-2x^{2}-yx[/tex]
    [tex]=xy-2x^{2}[/tex]

    [tex]D_{x}V=0:[/tex]
    [tex]0=xy-2x^{2}[/tex]
    [tex]2x^{2}=xy[/tex]
    [tex]2x=y[/tex]

    [tex]x^{2}+xy=300[/tex]
    [tex]x(x+y)=300[/tex]
    [tex]2x=y[/tex]
    [tex]x(x+2x)=x(3x)=300[/tex]
    [tex]x^{2}=100[/tex]
    [tex]x=10[/tex]

    x | 0 _ 10
    y | 300 0 20
    V | 0 0 2000

    Therefore x=10, y=20

    Chk [tex]D^{2}_{x}V<0:[/tex]
    [tex]D^{2}_{x}V=y+xD_{x}y-4x=y-(2+y/x)x-4x<0[/tex] when x=10, y=20
    => abs max
     
    Last edited: Oct 27, 2010
  2. jcsd
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