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6 April, 20:39

A cylindrical metal container, open at the top, is to have a capacity of 24 cu. in. The cost of material used for the bottom of the container is $0.15/sq. in., and the cost of the material used for the curved part is $0.05/sq. in. Find the dimensions that will minimize the cost of the material, and find the minimum cost.

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  1. 6 April, 20:44
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    Answer: Dimensions:

    X = radius of base = 2.76 in L = height = 1 in

    Minimum cost = 4.456 $

    Step-by-step explanation:

    We have:

    V 24 in³ V = πx²L where x radius of base of the cylinder and h the height of the cylinder.

    then L = V/πx² L = 24/πx²

    Area (x) = Base area + lateral area

    A (x) = πx² + 2πxL ⇒ A (x) = πx² + 2πx (24/πx²) ⇒A (x) = πx² + 48/x

    Taking derivatives:

    A' (x) = 2πx - 48/x² A' (x) = 0 6.28x² - 48 = 0

    x² = 48 / 6,28 x² = √7.64 x = 2.76 radius of base (in)

    and the height L = 24 / (3,14) * (2.76) ² L = 1 in

    cost : C ($) = (0,15) * (2.76) ²*3.14 + 2 * (3.14) (2.76) * 1 * 0.05

    C ($) = 3.59 + 0.866

    C ($) = 4.456 $
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