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A cardboard box manufacturer wishes to make open boxes from rectangular pieces of cardboard with dimensions 40 cm by 60 cm by cutting equal squares from the four corners and turning up the sides. Find the length of the side of the cut-out square so that the box has the largest possible volume. Also, find the volume of the box.
in Grade 12 Maths by Diamond (49,543 points) | 22 views

1 Answer

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Let each side of the cut-out squares be x cm

So:

Length of base of box = 60 - 2x

Width of base of box = 40 - 2x

Area of each cut-out square = x^2

Volume of box

    = (60 - 2x)(40 - 2x)(x) = (2400 - 120x - 80x + 4(x^2))(x)

    = 4(x^3) - 200(x^2) + 2400x

d(volume)/dx = 12(x^2) - 400x + 2400

For maximum volume, d(volume)/dx = 0

12(x^2) - 400x + 2400 = 0

3((x^2) - 100x + 600 = 0

Using quadratic formula, x = (100 ± √((100^2) -(4 x 3 x 600))) /(2 x 3)

x = (100 ± √2800)/6 = 25.5 or 7.85

But since width = 40 - 2x, this implies 0 < x < 20

So the feasible side of the cut-out squares is 7.85 cm, which is the length of the side of the cut-out square so that the box has the largest possible volume.

Volume of the box = 4(x^3) - 200(x^2) + 2400x

    = 4(7.85^3) - 200(7.85^2) + 2400(7.85)

    = 8450.45 cm3
by Silver Status (28,731 points)

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