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calculus

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A box is constructed out of two different types of metal. The metal for the top and bottom, which are both square, costs $2/ft2. The metal for the four rectangular sides costs $3/ft2. Find the dimensions that minimize cost if the box has a volume 20 ft3.

Top and bottom should be squares with sides of length L = ft
The height of the box should be h = ft

  • calculus - ,

    Cost = cost of top + cost of bottom + cost of 4 sides
    = 2L^2 + 2L^2 + 3(4Lh)
    = 4L^2 + 12Lh

    but L^2 h = 20
    h = 20/L^2

    so Cost = 4L^2 + 12L(20/L^2)
    = 4L^2 + 240/L
    d(Cost)/dL = 8L - 240/L^2 = 0 for max/min of Cost

    8L = 240/L^2
    L^3 = 30
    L = 3.107

    then h = 20/3.107^2 = 2.07

  • calculus - ,

    V-Volume
    V=20
    V=L^2*h
    L^2*h=20 Divide with (L^2)
    h=20/(L^2)
    Area of square is L^2
    Areao of rectacangle L*h
    Total price=2*Area of square*(2$)+4*Area of rectacangle*(3$)
    P-Price
    P=2*(L^2)*2+4*L*h*3
    =4*L^2+12*L*20/(L^2)
    =4*L^2+240*L/(L^2)
    =4*L^2+240/L

    (dP/dL)=4*2*L+240*(-1)/(L^2)
    (dP/dL)=8*L-240/(L^2)
    Price have minimum where is (dP/dL)=0

    8*L-240/(L^2)=0
    8*L=240/(L^2) Divide with 8
    L=30/(L^2) Multiply with L^2
    L^3=30

    L=third root of 30

    h=20/(L^2)
    =20/(third root of 30)^2

    h=20/third root of 900

    Proof that is minimum
    Function have minimum when:

    (dP/dL)=0 and (d^2P)/dL^2>0

    First derivation=0 and
    second derivation higher of zero

    (d^2P)/dL^2=Derivation of first derivation

    (d^2P)/dL^2=8-240*(-2)/L^3
    =8+480/L^3

    8+480/L^3 is always higher of zero.

    Function have minimum.

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