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Math Help - Pre-Calc

  1. #1
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    Pre-Calc

    A rectangle has a perimeter of 80 cm. If its width is x, express its length and its area in terms of x. What is the maximum area of the rectangle?
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  2. #2
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    Quote Originally Posted by ep78 View Post
    A rectangle has a perimeter of 80 cm. If its width is x, express its length and its area in terms of x. What is the maximum area of the rectangle?
    Let x be the width and y be the height. Thus 2x+2y=80. Solving for y, we get x+y=40 and y=40-x.

    Let A(x) be an area of function. A(x) = x(40-x). That's just the length times the width all in terms of x.

    Now differentiate and find the critical point.
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    Quote Originally Posted by Jameson View Post
    Let x be the width and y be the height. Thus 2x+2y=80. Solving for y, we get x+y=40 and y=40-x.

    Let A(x) be an area of function. A(x) = x(40-x). That's just the length times the width all in terms of x.

    Now differentiate and find the critical point.

    so would i plug in any number into x(40-x) or is there a certain one?
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  4. #4
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    Ooops! This is pre-calc. I was throwing in some Calculus terms.

    You could graph this area function and visually spot the maximum. I will say that a square is the best shape at maximizing area when the perimeter is constrained.
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  5. #5
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    oh haha, okay thank you very much!
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  6. #6
    Len
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    Quote Originally Posted by ep78 View Post
    so would i plug in any number into x(40-x) or is there a certain one?
    Differentiate then find the critical numbers. (Where the derivative = 0)

    dy/dx=40-2x

    40-2x=0

    x=20
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    Quote Originally Posted by Len View Post
    Differentiate then find the critical numbers. (Where the derivative = 0)

    dy/dx=40-2x

    40-2x=0

    x=20
    Look at title. I made the same mistake. (Pre-calc)
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  8. #8
    Len
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    Quote Originally Posted by Jameson View Post
    Look at title. I made the same mistake. (Pre-calc)
    Ahh sorry my internet crashed when I was replying and when it came back online I pressed refresh and it sent.
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