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MarcLeclair

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y=5-x^2 . What are the dimensions of such a rectangle with the greatest possible area?

  • one year ago
  • one year ago

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  1. MarcLeclair
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    @zepdrix

    • one year ago
  2. MarcLeclair
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    I can't seem to find the the constraint and the good equation to make the problem work :/ except for x=-y - y

    • one year ago
  3. MarcLeclair
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    or wait forget that last equation :P

    • one year ago
  4. zepdrix
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    |dw:1355978532328:dw|Let's see if we can get the box drawn accurately.

    • one year ago
  5. MarcLeclair
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    yeah I have it drawn but im still wondering if its symmetrical :P

    • one year ago
  6. zepdrix
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    |dw:1355978702013:dw|Understand how I labeled those lengths?

    • one year ago
  7. zepdrix
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    Yes it's symmetrical, both corners have to touch the parabola. And in order for the top to be a straight line across, the points have to touch at the same places across the parabola.

    • one year ago
  8. MarcLeclair
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    yeah it make sense, but shouldnt it be 1/2 x ? or its just 2 rectangles then?

    • one year ago
  9. zepdrix
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    It's one large rectangle, touching one end of the parabola and the other end as well. We're basically choosing a point along the parabola. The coordinates that correspond to that point we call (x,y)|dw:1355978868031:dw|And due to symmetry, it'll touch at the same spot on the other side.

    • one year ago
  10. MarcLeclair
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    yeah it makes sense, the y should be the same on both side, just the roots will be different but the x is the same but on the negative side xD

    • one year ago
  11. zepdrix
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    Yah :) So if we want a function for AREA. We need to establish the LENGTH and WIDTH of this rectangle. The Width is pretty clearly Y. What about the Length? Remember, when dealing with length, we don't care about the negative. We can't have a wooden board that is -2 inches long.

    • one year ago
  12. MarcLeclair
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    yeah so it would be 2x?. However I'll stop you for a sec, everytime I do an optimization i set myself a constraint but there's none here.. am i wrong?

    • one year ago
  13. zepdrix
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    The constraint is the parabola, we are constrained to the shape of the parabola. Otherwise the box could just grow and grow and grow.

    • one year ago
  14. MarcLeclair
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    so the constraint is already labelled as y=5-x^2! smart :P

    • one year ago
  15. zepdrix
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    Ah yes, that's convenient :)

    • one year ago
  16. MarcLeclair
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    so then, if that is my constraint a= 2x(y) and I can replace y with 5-x^2?

    • one year ago
  17. zepdrix
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    \[\huge A=2x\color{orangered}{y}\]Yes very good :)

    • one year ago
  18. MarcLeclair
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    so then i find my derivative as a'(x) = 6x^2 -10x. I get my 0 as +/- sqrt(10/6), do I need to x2 for my full width?

    • one year ago
  19. zepdrix
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    Woops, I think your 10x will turn into 10 when you differentiate it, right?

    • one year ago
  20. MarcLeclair
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    my answer is still wrong x.x and yes it does , hahah

    • one year ago
  21. MarcLeclair
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    nvm my answer is 2sqrt(10/6), but I have to multiply by 2 right?

    • one year ago
  22. zepdrix
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    \[\large A=10x-2x^3, \qquad A'=10-6x^2\]\[\large x=\pm \sqrt{\frac{5}{3}}\]So you're getting a critical point here, yes? Simplify your fraction :D heh

    • one year ago
  23. MarcLeclair
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    thanks a lot youre REALLLY helpful, i might pass my final thanks to you haha

    • one year ago
  24. zepdrix
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    We only care about the positive square root (since it's a length). So I guess I didn't need to worry about the p/m sign. :)

    • one year ago
  25. zepdrix
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    So now what we want to do is ummm

    • one year ago
  26. zepdrix
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    Plug our critical point into our constraint, to find a corresponding length for Y.

    • one year ago
  27. MarcLeclair
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    we multiply by 2 because one x is only half the width, than plug it in y = 5-x^2 and were done :D

    • one year ago
  28. zepdrix
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    Bah yer too fast! lol

    • one year ago
  29. zepdrix
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    So in this problem, they don't want the MAXIMUM area, they just want the LENGTH and WIDTH of the box. We're already half done. We found x. Or rather as you pointed out, 2x. Now we just need Y.

    • one year ago
  30. MarcLeclair
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    Which i found as 20/6

    • one year ago
  31. MarcLeclair
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    :) thanks a lot mate!!!

    • one year ago
  32. zepdrix
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    Yay team! :) No problem.

    • one year ago
  33. zepdrix
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    But again, simplify that ugly fraction! lol 20/6 = 10/3

    • one year ago
  34. MarcLeclair
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    still ugly in my world. I don't know how you do it but I can never see through this small number of information im given. Or I barely can hahaha. Thanks!

    • one year ago
  35. zepdrix
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    \[\large Length=2\sqrt{\frac{5}{3}}, \qquad Width=\frac{10}{3}\]

    • one year ago
  36. zepdrix
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    Yah related rates are tough to get a grip on. They show up in many many different forms. :)

    • one year ago
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