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MoreForJoe Group Title

minimizing area: A 36-in. piece of wire is cut into two pieces. One Piece is used to form a circle while the other is used to form a square. How should the wire be cut so that the sum of the areas is minimal? What is the minimum value?

  • 2 years ago
  • 2 years ago

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  1. Dalvoron Group Title
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    |dw:1323260051758:dw| The sum of the areas is\[A(r)=\frac{1}{2}(2\pi)r^2 + (36-2\pi r)^2\]You want this to be a \(\textit{minimum}\). This will happen when the first derivative of A = 0. (though this could also be a maximum. To figure out which, you'd need the second derivative I think).

    • 2 years ago
  2. MoreForJoe Group Title
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    what do i do next?

    • 2 years ago
  3. Dalvoron Group Title
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    That should get you a value for r.

    • 2 years ago
  4. Dalvoron Group Title
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    Then you work out the length \(2\pi r\), and the length \((36 - 2\pi r)\)

    • 2 years ago
  5. Dalvoron Group Title
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    Actually, according to my sanity test, that gives you lengths longer than 36, so it can't be right.

    • 2 years ago
  6. Dalvoron Group Title
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    Nope, calculation error. Never mind.

    • 2 years ago
  7. MoreForJoe Group Title
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    i never did one of these problems and im sure he's going to put it on the test so i have to know how to do it -..-

    • 2 years ago
  8. MoreForJoe Group Title
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    i still dont know where to begin...

    • 2 years ago
  9. Dalvoron Group Title
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    The total length of the wire is 36 in, right? That length will then turn into two lengths when you cut it in half. One length will become the circumference of your circle (\(2\pi r)\), and the other will become the perimeter of your square (36 - \(2\pi r\)). So now you have: |dw:1323260762343:dw| I actually made a mistake earlier, I forgot the perimeter of the square and the length of its side are not the same. So the area of the circle is \(\pi r^2\), and the area of the square is the side length squared. The side length is one quarter of the perimeter, so the area of the square is \(\left(\frac{1}{4}(36-2\pi r)\right)^2\) Thus, the sum of the two areas is....?

    • 2 years ago
  10. MoreForJoe Group Title
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    πr^2+((1/4)(36-2πr))^2?

    • 2 years ago
  11. Dalvoron Group Title
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    Precisely. So \[A(r)=\pi r^2 + 81-9\pi r+\frac{1}{4}\pi^2r^2\]when you simplify, and expand that bracket. What we have is a function, which defines the area in terms of the radius. To give you an idea of what that looks like, go to http://www.wolframalpha.com/input/?i=%5Cpi+r^2%2B%289-%5Cfrac{%5Cpi}{2}r%29^2 We're looking for that lowest point on the graph. Where is that point? Well it's at the point where the slope of the graph is zero. The slope of the graph is given by \[\frac{dA}{dr}\], and we want it to be equal to zero, so we say \[\frac{dA}{dr}=0\]

    • 2 years ago
  12. MoreForJoe Group Title
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    where did dA/dr come from?

    • 2 years ago
  13. Dalvoron Group Title
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    Have you started calculus yet?

    • 2 years ago
  14. MoreForJoe Group Title
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    no...

    • 2 years ago
  15. Dalvoron Group Title
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    Good grief. I have no idea how to solve this problem without calculus. You've not seen anything like dy/dx yet?

    • 2 years ago
  16. MoreForJoe Group Title
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    no... my class is intermediate algebra

    • 2 years ago
  17. Dalvoron Group Title
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    Well my apologies, I don't see how it's possible to solve this with just algebra.

    • 2 years ago
  18. MoreForJoe Group Title
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    hmm... can you type how to solve this your way and i can just write it down?

    • 2 years ago
  19. Dalvoron Group Title
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    It's pretty advanced mathematics compared to what you're currently doing, there's just no way you'd be expected to do it this way.

    • 2 years ago
  20. MoreForJoe Group Title
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    true...

    • 2 years ago
  21. alexxis15 Group Title
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    WELL THAT'S EASY

    • 2 years ago
  22. MoreForJoe Group Title
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    huh?

    • 2 years ago
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