Anonymous
Anonymous asked in Science & MathematicsMathematics · 11 months ago

Calculus: The length plus the girth of any piece of airline luggage length must be no more than 158 cm.?

Determine the dimensions of a cylindrical package with maximum volume that would be meet this requirement.

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  • 11 months ago
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    The dimensions we need to worry about here is the diameter of the base and the height.

    This value cannot be more than 158 cm. Let's set up this equation:

    158 = d + h

    In terms of radius:

    158 = 2r + h

    Let's solve this for r in terms of h:

    158 - h = 2r

    79 - h/2 = r

    We want a maximum volume of the cylinder, so let's get that equation:

    V = πr²h

    If we substitute the expression above into this equation we will get volume in terms of only the height:

    V = π(79 - h/2)²h

    Simplify:

    V = π(6241 - 79h + h²/4)h

    V = 6241πh - 79πh² + (π/4)h³

    The maximum volume happens when the first derivative is zero, so:

    dV/dh = 6241π - 158πh + (3π/4)h²

    0 = 6241π - 158πh + (3π/4)h²

    Divide both sides by π, then put into standard polynomial form:

    0 = 6241 - 158h + (3/4)h²

    0 = (3/4)h² - 158h + 6241

    Let's simplify further by multiplying both sides by 4:

    0 = 3h² - 632h + 24964

    Now we can solve for this quadratic and throw out any values that don't make sense:

    h = [ -b ± √(b² - 4ac)] / (2a)

    h = [ -(-632) ± √((-632)² - 4(3)(24964))] / (2 * 3)

    h = [ 632 ± √(399424 - 299568)] / 6

    h = [ 632 ± √(99856)] / 6

    h = (632 ± 316) / 6

    h = 316 / 6 and 948 / 6

    h = 158/3 and 158

    Since a height of 158 would result in a radius of 0, a volume of 0, this is a minimum so we'll throw that out.

    h = 158/3

    Now that we have h, solve for r:

    r = 79 - h/2

    r = 79 - (158/3)/2

    r = 79 - 158/6

    r = 474/6 - 158/6

    r = 316/6

    r = 158/3

    Solving for d now:

    d = 2r

    d = 2(158/3)

    d = 316/3

    The maximum volume would have a diameter of 316/3 (105.33) cm and a height of 158/3 (52.67) cm

    The sum of this diameter and height will be 158 cm:

    105.33 + 52.67 = 158

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    • david h
      Lv 6
      11 months agoReport

      Also you have not used any calculus to solve the problem and the question states Calculus problem in its title.

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  • Ian H
    Lv 7
    11 months ago

    Girth is not the same as diameter. It refers to the distance around the outside of a thick or fat object, like a strap around a tree or the waist of a person.

    In this example it is the circumference of the cylinder, so the condition is

    2πr + h = 158

    V = πr^2h = πr^2*(158 - 2πr) = 2π(79r^2 - πr^3)

    dV/dr = 2π(158r - 3πr^2)

    Maximum when r = 158/(3π) ~ 16.76, and,

    h = 158 - 2πr = 158/3 ~ 52.67

    Note that h = πr for max volume, so that 2πr + h = 3πr = 158

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  • Mike G
    Lv 7
    11 months ago

    Girth = 2πr

    h = Length = 158-2πr

    V = volume = 2πr(158-2πr)

    V = 316πr-4π^2r^2

    dV/dr = 316π - 8π^2*r = 0

    316 - 8π*r = 0

    r = 316/(8π) = 12.573 cm

    h = 79 cm

    • david h
      Lv 6
      11 months agoReport

      Well done ! you seem to be the only person who got the correct answer. I like your solution , nice and clear.

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  • Dixon
    Lv 7
    11 months ago

    V = πr²h

    V = (π) rrh

    If we compare this to a cube V = s³ we can see π is a scaling factor to account for the shape of the cross section.

    And when the factor of π is isolated we can see that rrh compares to sss

    and the cylinder will thus be maximal at h = r,

    giving (π) rrr, where each r effectively represents an orthogonal direction.

    Since girth is from two equally weighted orthogonal directions and length is the third,

    the sum of girth and length weights each direction the same. Then...

    r = 158/3, h = 158/3

    • ...Show all comments
    • Dixon
      Lv 7
      11 months agoReport

      i know girth is the circumference but i did goof, it should be h = 158/3 like u got, then girth = 2 x 158/3

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  • 11 months ago

    v = pi * r^2 * h

    d + h = 158

    d = 2r

    2r + h = 158

    h = 158 - 2r

    v = pi * r^2 * (158 - 2r)

    v = 158*pi*r^2 - 2*pi*r^3

    dv/dr = 316*pi*r - 6*pi*r^2

    To maximize v, set the derivative to zero:

    316*pi*r - 6*pi*r^2 = 0

    pi*r*(316 - 6r) = 0

    r = 0 or 6r = 316

    r = 0 or r = 316/6 = 158/3 =~ 52.66666...

    h = 158 - 2(158/3) = 158/3 =~ 52.66666...

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  • GTB
    Lv 7
    11 months ago

    V = (Pi)r^2h; h + (Pi)r^2 = 158 so (Pi)r^2 = 158-h; therefore V = (158-h)h and we want to maximize V; take dV/dh, set it = 0, solve for h; dV/dh = 158-h = 0; h = 79 cm; solve for r knowing that (Pi)r^2 = 158-h or (Pi)r^2 = 158-h = 79 or r = [(79/(Pi)]^0.5

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