Find the dimensions of a rectangle with area 512 m^2 whose perimeter is as small as possible. ?

(If both values are the same number, enter it into both blanks.)

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  • 2 weeks ago

    A rectangle with an area of 512 m^2

    whose perimeter is as small as possible is a

    a square which measures 16√2 m on a side (approx 22.627 m)

  • RR
    Lv 7
    2 weeks ago

    It has to be a square.

    Try it with 16 square m

    it could be 4 x 4 (perimeter 4 + 4 + 4 + 4 = 16)

    It could be 2 x 8 (perimeter 2 + 2 + 8 + 8 = 20)

    it could be 1 x 16 (perimeter 1 + 1 + 16 + 16 = 34)

    The 4 x 4 square has the smallest perimeter.

    So all you need to do is find the square root of 512

    sq rt 512 = 22.627M

    It's a square 22.627 x 22.627

  • ?
    Lv 7
    2 weeks ago

    Area A = ℓ x w = 512 m² ⇒ ℓ = (512 m²)/w

    Perimeter P(w) = 2ℓ + 2w = 2(512 m²)/w + 2w

    ................P(w) = 1024/w + 2w

    ...........................1024 + 2w²

    ................P(w) = --------------, w > 0

    ..................................w

    // We have the restriction w > 0 for P(w)

    // because you can't have w=0 in the denominator

    // and you can't have a negative length

    // Find the 1st derivative P '(w)

    .............w [4w] - (1024 + 2w²)[1]

    P '(w) = --------------------------------

    ..........................w²

    .............4w² - 1024 - 2w²

    P '(w) = ----------------------

    ..........................w²

    .............2w² - 1024

    P '(w) = ---------------

    ...................w²

    // Find the critical values, where P '(w) = 0

    P '(w) = 0 ⇔ 2w² - 1024 = 0

    ................⇒ 2w² = 1024

    ................⇒ w² = 512

    ................⇒ w = ± 8√2

    // Find the 2nd derivative P "(x)

    ..............w² [4w] - (2w² - 1024)[2w]

    P "(w)  = ----------------------------------

    .........................(w²)²

    ..............2048w

    P "(w) = ----------

    ................w⁴

    // Apply the 2nd derivative test to the critical values.

    When w = -8√2,

    P "-(8√2) < 0 ⇒ P(w) has a relative MAX at w= -8√2

    When w = +8√2

    P "(8√2) > 0  ⇒ P(w) has a relative MIN at w = +8√2

    // Therefore, the dimensions that result in a minimum perimeter are

    w = +8√2 and

    ℓ = 512/8√2 = 32√2................ANS

  • 2 weeks ago

    A square is the rectangle whose perimeter is the smallest for a given area.

    l = w = √512

    = √256√2

    = 16√2

    ≈ 22.627417 m

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  • Amy
    Lv 7
    2 weeks ago

    xy = 512

    p = 2x + 2y

    p = 2x + 1024/x

    Solve dp/dx= 0

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