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MarcLeclair

  • 2 years ago

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?

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  1. MarcLeclair
    • 2 years ago
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    @zepdrix

  2. MarcLeclair
    • 2 years ago
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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

  3. MarcLeclair
    • 2 years ago
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    or wait forget that last equation :P

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

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

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

  7. zepdrix
    • 2 years ago
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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.

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

  9. zepdrix
    • 2 years ago
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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.

  10. MarcLeclair
    • 2 years ago
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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

  11. zepdrix
    • 2 years ago
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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.

  12. MarcLeclair
    • 2 years ago
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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?

  13. zepdrix
    • 2 years ago
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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.

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

  15. zepdrix
    • 2 years ago
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    Ah yes, that's convenient :)

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

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

  18. MarcLeclair
    • 2 years ago
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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?

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

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

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

  22. zepdrix
    • 2 years ago
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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

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

  24. zepdrix
    • 2 years ago
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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. :)

  25. zepdrix
    • 2 years ago
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    So now what we want to do is ummm

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

  27. MarcLeclair
    • 2 years ago
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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

  28. zepdrix
    • 2 years ago
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    Bah yer too fast! lol

  29. zepdrix
    • 2 years ago
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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.

  30. MarcLeclair
    • 2 years ago
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    Which i found as 20/6

  31. MarcLeclair
    • 2 years ago
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    :) thanks a lot mate!!!

  32. zepdrix
    • 2 years ago
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    Yay team! :) No problem.

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

  34. MarcLeclair
    • 2 years ago
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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!

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

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

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