Maximum Area

Discussion in 'Other Pre-University Math' started by nycmathguy, Aug 25, 2021.

  1. nycmathguy

    nycmathguy

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    Section 2.1
    Question 73

    For this question, I seek a hint for parts a & b.

    20210825_020636.jpg
     
    nycmathguy, Aug 25, 2021
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  2. nycmathguy

    MathLover1

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    a hint for parts a : the length of corrals is 2x, the width is y
    fencing or perimeter=200ft

    & b: to determine the maximum first find x and y
     
    MathLover1, Aug 25, 2021
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  3. nycmathguy

    nycmathguy

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    Thanks for the hints. Will work on this and show my work.
     
    nycmathguy, Aug 26, 2021
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  4. nycmathguy

    nycmathguy

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    Part (a)

    P = 200

    200 = 2(2x) + 2y

    200 = 4x + 2y

    200 = 2(2x + y)

    200/2 = 2x + y

    100 = 2x + y

    (100 - 2x) = y

    A(x) = length •width

    A(x) = 2x (100 - 2x)

    A(x) = 200x - 4x^2

    Tell me if this right before moving on to part (b).
     
    nycmathguy, Aug 26, 2021
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  5. nycmathguy

    MathLover1

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    correct
     
    MathLover1, Aug 26, 2021
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  6. nycmathguy

    nycmathguy

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    I will do part (b) tomorrow.
     
    nycmathguy, Aug 26, 2021
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  7. nycmathguy

    nycmathguy

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    A(x) = 200x - 4x^2

    Let A(x) = 0

    0 = 200x - 4x^2

    I found x to be 0 and 50. Since we are dealing with measurements in terms of feet, I rejected x = 0.

    Now that I found x to be 50, where do I plug my x-value to find y?
     
    nycmathguy, Aug 27, 2021
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  8. nycmathguy

    MathLover1

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    in part a
    200 = 2(2x) + 2y
     
    MathLover1, Aug 28, 2021
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  9. nycmathguy

    nycmathguy

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    Let x = 50

    200 = 2(2x) + 2y

    200 = 2(2•50) + 2y

    200 = 2(100) + 2y

    200 = 200 + 2y

    200 - 200 = 2y

    0 = 2y

    0/2 = y

    0 = y

    What does x = 50 & y = 0 represent here?
    Has part (b) been answered?
     
    nycmathguy, Aug 28, 2021
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  10. nycmathguy

    MathLover1

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    I gave you bad hint, a hint for part a : the perimeter=200ft includes the fence in the middle, the shared fence

    so you have perimeter

    Let x represent the dimension perpendicular to the shared fence. (these 2x)
    Then the dimension parallel to the shared fence will be: (3y)
    3y = 200-2x
    y = (200-2x)/3
    y = (2/3)(100-x)

    The area will be the product of these dimensions, so will be:

    area = x*y
    area = x(2/3)(100-x)
    area = (2/3)(100x -x^2)
    area = (-2/3)(x^2 -100x)

    To complete the square, we need to add the square of half the x-coefficient inside parentheses, and the opposite of that quantity outside parentheses.

    area = (-2/3)(x^2 -100x+50^2)-(-2/3)50^2. . . . complete the square

    area = (-2/3)(x -50)^2 +5000/3

    The vertex of this parabolic curve is at x=50, so the dimensions of the maximum area is at x = 50
    then substitute it in
    y = (200 -2*x)/3

    y = (200 -2*50)/3
    y=100/3
    y= 33 &1/3
    feet
     
    MathLover1, Aug 28, 2021
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  11. nycmathguy

    nycmathguy

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    Great reply. Don't worry about giving me the wrong hint. It happens to everyone. More math later from the beach.
     
    nycmathguy, Aug 28, 2021
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  12. nycmathguy

    MathLover1

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    I just don't know why question was to express the area in terms of x when there is much faster and easier way to get same answer

    given perimeter 200ft
    half of it is 2 times the length 2x=>means 2*2x=100=>the length 2x=50ft
    half of it is 3 times the width 3y=>means 3y=100=>the width y=100/3 =33&1/3ft=33.33ft

    then area=2x*y=50ft*33.33ft=1666.5ft^2
     
    MathLover1, Aug 28, 2021
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  13. nycmathguy

    nycmathguy

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    I like the shortcut you provided.
     
    nycmathguy, Aug 28, 2021
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