Optimization Problem?

Find the dimensions of the rectangle of maximum area that can be formed from a 50-in piece of wire.

What are the dimensions and maximum area?

3 Answers

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  • Philip
    Lv 6
    2 months ago

    Let (length,width) of rectangle = (x,y).

    Perimeter of rectangle = 2(x+y) = 50 and y = 25-x.

    Area of rectangle = f(x) = x(25-x) = 25x - x^2.

    f'(x) = 25 - 2x.

    f''(x) = -2.

    An extremum occurs wherever f'(x) = 0.

    f'(x) = 0 for x = (25/2).

    f''(x) = -2, < 0, ---> extremum is a maximum.

    f(25/2) = (25/2)[2(25/2) -1(25/2)] = (25/2)^2 = (625/4) = 156.25.

    The rectangle is actually a square of side length (25/2).

    • ...Show all comments
    • Philip
      Lv 6
      2 months agoReport

      Sadly, I must reject your criticism until you present valid proof that I did, in fact, make several
      mistakes. Where your thumbs point is of little concern to me. 

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  • ted s
    Lv 7
    2 months ago

    it is well known that the figure is a square

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  • rotchm
    Lv 7
    2 months ago

    Let x & y be the width & length resp.

    You Are given that 2x+2y = 50, right?

    And the area is A = xy, right?

    From the first eqs, solve for y. So you have y = y(x). Plug this y(x) into the second eqs. You now have the area A(x) solely as a fct of x. What is this function A(x) ?

    Answer that & we will proceed. 

    • Philip
      Lv 6
      2 months agoReport

      Congratulations. I did exactly that which you did and proceeded without your permission. I felt,
      unlike you, no need to explain length & width & how area of a rectangle is calculated as if my fellow participants were totally ignorant children. 

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