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28 September, 01:47

The top and bottom margins of a poster are 6 cm and the side margins are each 2 cm. If the area of printed material on the poster is fixed at 390 square centimeters, find the dimensions of the poster with the smallest area.

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  1. 28 September, 02:39
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    A (min) = 711,60 cm²

    L = 15,40 cm

    D = 46,21 cm

    Step-by-step explanation:

    Let L and D the lenght and height of the poster

    and x and y dimensions of printed area so

    L = y + 12 and D = x + 4

    Printed area = 390 cm² and Pa = x*y then y = 390/x

    Poster area is Ap = L * D = (y + 12) * (x + 4)

    A (x) = (390/x + 12) * (x + 4)

    A (x) = 390 + 1560/x + 12x + 48

    A (x) = 438 + 1560/x + 12x

    Taking derivatives on both sides of the equation

    A' (x) = 12 - 1560/x²

    A' (x) = 0 12 - 1560/x² = 0 12x² = 1560

    x² = √ 130

    x = 11.40 cm then y 390 / 11,40 y = 34,21 cm

    And L = 11,40 + 4 L = 15,40 cm

    And D = 34,21 + 12 D = 46,21 cm

    A (min) = 15,40 * 46,21

    A (min) = 711,60 cm²
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