Calculus

A box (with no top) is to be constructed from a piece of cardboard of sides A and B by cutting out squares of length h from the corners and folding up the sides. Find the value of h that maximizes the volume of the box if
A = 7 and B = 12

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  1. Draw a rectangle 7 x 12.

    In the corners draw four squares of dimensions h x h.

    It doesn't matter what the length of h is.

    Dimensions of a new rectangle will be ( 7 - 2 h ) x ( 12 - 2 h )

    The volume of the box will be:

    Area of a new rectangle ∙ length h

    V = ( 7 - 2 h ) ∙ ( 12 - h ) ∙ h

    V = ( 84 - 24 h - 7 h + 2 h² ) ∙ h

    V = ( 2 h² - 31 h + 84 ) ∙ h

    V = 2 h³ - 31 h² + 84 h

    The function has an exstrem (maximum or minimum) where the first derivative is zero.

    In this case you need to find:

    dV / dh = V′( h ) = 0

    ( 2 h³ - 31 h² + 84 h )′ = 2 ∙ 3 h² - 31 ∙ 2 h + 84 = 0

    6 h² - 62 h + 84 = 0

    The solutios are:

    h = ( 31 -√457 ) / 6 ≈ 1.60374

    and

    h = ( 31 +√457 ) / 6 ≈ 8.7296

    Now you have to second derivative test.

    If V"( h ) < 0 then function has maximum at h.

    If V" ( h ) > 0 then function has minimum at h.

    V" ( h ) = 2 ∙ 6 h - 62

    V" ( h ) = 12 h - 62

    For

    h = 1.60374

    V" ( h ) = 12 ∙ 1.60374 - 62 = − 42​.75512 < 0

    For

    h = 8.7296

    V" ( h ) = 12 ∙ 8.7296 - 62 = 42.7552 > 0

    So

    For h = ( 31 -√457 ) / 6 ≈ 1.60374 volume has a maximum.

    For h = ( 31 +√457 ) / 6 ≈ 8.7296 volume has a minimum.

    By the way dimension h = 8.7296 are impossible because for h = 8.7296 dimension 7 - 2 h would have a negative value and the length cannot be negative.

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