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29 July, 10:24

The base and sides of a container is made of wood panels. The container does not have a lid. The base and sides are rectangular. The width of the container is x c m. The length is fourth times the width. The volume of the container is 400 c m 3. Determine the minimum surface area that this container will have.

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  1. 29 July, 10:52
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    A (mim) = 300 cm²

    Step-by-step explanation:

    Area of the base is:

    A = 4*x*x ⇒ A = 4*x²

    Let call h the height of the container, then area lateral are:

    A₁ = 2*4*x*h A₁ = 8*x*h and

    A₂ = 2 * x * h A₂ = 2*x*h

    From the volume of the container we have:

    400 = Area of the base*h

    400 = 4*x²*h ⇒ h = 100 / x²

    Now Total area of the container is:

    A = 4*x² + 8*x*h + 2*x*h ⇒ A = 4*x² + 10*x*h

    As h = 100/x²

    A (x) = 4*x² + 10*x * 100/x²

    A (x) = 4*x² + 1000/x

    Taking derivatives on both sides of the equation we get:

    A' (x) = 8*x - 1000/x²

    A' (x) = 0 ⇒ 8*x - 1000/x² = 0

    8*x³ = 1000 = 0 ⇒ x³ = 1000 / 8 ⇒ x = ∛ (1000) / 8

    x = 5 cm

    Now minimum area is:

    Area of the base

    4*5*5 = 100 cm²

    A₁ = 8*x*h h = 100 / x² h = 100 / 25 h = 4 cm

    A₁ = 8*5*4

    A₁ = 160 cm²

    A₂ = 2*5*4

    A₂ = 40 cm²

    A (mim) = 300 cm²
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