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The dimensions of a gift box are consecutive positive integers such that the height is the least integers and the length is the greatest integer. If the height is increased by 1 cm, the width is increased by 2 cm, and the length is increased by 3 cm, then the larger box is constructed such that the volume is increased by 456 cm3. Determine the dimensions of the original box.

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  • 2 months ago
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    You are told that a box has dimensions consisting of positive consecutive integers where the height is the least and the length is the most.

    So if we call the width "x", then the others are:

    h = x - 1

    w = x

    l = x + 1

    If the height is increased by 1 (now x)

    and the width is increased by 2 (now x + 2)

    and the length increased by 3 (now x + 4)

    the volume is increased by 456 cm³

    So let's get expressions for the volumes of both boxes:

    V = lwh

    V₁ = (x + 1)x(x - 1) and V₂ = (x + 4)(x + 2)x

    And we know that:

    V₂ - 456 = V₁

    So this gives us a system of three equations and three unknowns that we can solve for.

    Let's first substitute the expression in terms of V₂ for V₁.  Then to simplify things we'll remove the subscript from the V's as we don't won't need them anymore:

    V₁ = (x + 1)x(x - 1) and V₂ = (x + 4)(x + 2)x

    V₂ - 456 = (x + 1)x(x - 1) and V₂ = (x + 4)(x + 2)x

    V - 456 = (x + 1)x(x - 1) and V = (x + 4)(x + 2)x

    Now let's susbtitute the expression in terms of x in for V in the first equation to have one equation with one unknown to solve:

    V - 456 = (x + 1)x(x - 1)

    (x + 4)(x + 2)x - 456 = (x + 1)x(x - 1)

    Let's expand and simplify both sides to see where we are at:

    (x² + 6x + 8)x - 456 = (x² - 1)x

    x³ + 6x² + 8x - 456 = x³ - x

    6x² + 9x - 456 = 0

    We can simplify both sides by dividing 3:

    2x² + 3x - 152 = 0

    Now I'll use quadratic equation:

    x = [ -b ± √(b² - 4ac)] / (2a)

    x = [ -3 ± √(3² - 4(2)(-152))] / (2 * 2)

    x = [ -3 ± √(9 + 1216)] / 4

    x = [ -3 ± √(1225)] / 4

    x = (-3 ± 35) / 4

    x = -38/4 and 32/4

    x = -19/2 and 8

    Since we were told that x must be a positive integer we can throw out the other:

    x = 8

    Now we can solve for l, w, and h:

    h = x - 1 and w = x and l = x + 1h = 8 - 1 and w = 8 and l = 8 + 1

    h = 7 and w = 8 and l = 9

    The original box had dimensions of 7 cm by 8 cm by 9 cm.

  • 2 months ago

    V = n (n+1)(n+2) 

    V' = (n+1)(n + 1 + 2)(n+2+3) = V + 456

    V = n(n^2 + 3n + 2) = n^3 + 3n^2 + 2n

    V' = (n+1)(n+3)(n+5) = 

    V' = (n+1)(n^2 + 8n + 15) = n^3 + 8n^2 + 15n + n^2 + 8n + 15 = n^3 + 9n^2 + 23n + 15 

    Hence 

    n^3 + 9n^2 + 23n + 15 = n^3 + 3n^2 + 2n + 456 

    'n^3' add out 

    Hence 

    6n^2 + 21n - 441 = 0 

    Factor out '3' 

    2n^2 + 7n - 147 = 0 

    Use Quadratic Eq'n 

    n = { -7 +/-sqrt[7^2 - 4(2)(-147)]} / 2(2) 

    n = { - 7 +/- sqrt[49 + 1176]} / 4

    n = { - 7 +/- sqrt[1225]} 4 

    n = { -7 +/- 35} / 4 

    n = 28/4 = 7  

    Hence the other dimensions are 8, & 9. 

    Verification 

    7 x 8 x 9 = 504

    8 x 10 x 12 = 960 

    960  - 504 = 456 as required.!!!!!

  • h = h

    w = h + 1

    l = w + 1 = h + 2

    h * (h + 1) * (h + 2) = v

    (h + 1) * (h + 1 + 2) * (h + 2 + 3) = v + 456

    (h + 1) * (h + 3) * (h + 5) = h * (h + 1) * (h + 2) + 456

    (h + 1) * (h + 3) * (h + 5) - h * (h + 1) * (h + 2) = 456

    (h + 1) * ((h + 3) * (h + 5) - h * (h + 2)) = 456

    (h + 1) * (h^2 + 8h + 15 - h^2 - 2h) = 456

    (h + 1) * (6h + 15) = 456

    (h + 1) * 3 * (2h + 5) = 456

    (2h + 5) * (h + 1) = 152

    2h^2 + 5h + 2h + 5 = 152

    2h^2 + 7h - 147 = 0

    h = (-7 +/- sqrt(49 + 4 * 2 * 147)) / 4

    h = (-7 +/- sqrt(49 + 8 * 3 * 49)) / 4

    h = (-7 +/- sqrt(49 * (1 + 24))) / 4

    h = (-7 +/- 7 * sqrt(25)) / 4

    h = -7 * (1 +/- 5) / 4

    h = -7 * (-4/4) , -7 * (6/4)

    h = -7 * (-1) , -7 * (3/2)

    h = 7 , -21/2

    h > 0

    h = 7

    w = 8

    l = 9

    7 * 8 * 9 = 504

    504 + 456 = 960

    7 + 1 = 8

    8 + 2 = 10

    9 + 3 = 12

    8 * 10 * 12 = 960

  • 2 months ago

    call the original dimensions x, y, z in order

    original volume V₀ = xyz

    new volume V₁ = (x+1)(y+2)(z+3) 

    still have 3 variables, trial and error next

    consecutive means choices are:

    x y z    V₀    x+1   y+2    z+3      V₁       Δ

    1 2 3     6      2       4        6       48       

    2 3 4   24      3       5        7      105      

    3 4 5   60      4       6        8      192

    4 5 6   120    5       7        9      315

    5 6 7   210    6       8       10     480

    6 7 8   336    7       9       11     693      357

    7 8 9   504    8      10      12     960      456

    ans: 7 8 9

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  • ?
    Lv 7
    2 months ago

    Let the dimensions of the box be L, W and H.

    (L + 3)(W + 2)(H + 1) - LWH = 456

    But W = L - 1 and H = L - 2

    (L + 3)(L - 1 + 2)(L - 2 + 1) - L(L - 1)(L - 2) = 456

    (L + 3)(L + 1)(L - 1) -L(L - 1)(L - 2) = 456

    Can you proceed from here ?

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