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imqwerty

  • one year ago

Hi ppl

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  1. Study_together
    • one year ago
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    What's your question?

  2. anonymous
    • one year ago
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    Hi ^-^ have a question ? :)

  3. imqwerty
    • one year ago
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    Wait m sending it

  4. Study_together
    • one year ago
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    ^_^

  5. imqwerty
    • one year ago
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    Jst one min

  6. imqwerty
    • one year ago
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  7. imqwerty
    • one year ago
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    @ganeshie8 @Study_together @Moe95

  8. Jadeishere
    • one year ago
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    I think you mean *People

  9. imqwerty
    • one year ago
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    yes u r correct @Jadeishere

  10. anonymous
    • one year ago
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    Do P and Q have to be adjacent?

  11. imqwerty
    • one year ago
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    yes P and Q have to be adjacent

  12. anonymous
    • one year ago
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    Let AC be represented by \(y_1=\frac43x_1\) and let CB be represented by \(y_2=-\frac34x_2+\frac{15}{4}\) on the specified intervals. (I skipped a few of the geometry steps, but basically, it looks like:) |dw:1437066809049:dw| We want to minimize the quantity \(y_1^2+y_2^2\) since that represents the sum of the areas. Our constraint is that the distance on the x-axis must correspond to 5 since it's a right triangle. So, I set up a constraint function. \[x_1+y_1+y_2+(5-x_2)=5\]\[\implies28x_1-21x_2=-45\]\[\implies x_1=\frac{21x_2-45}{28}\]Substituting in, we get \(y_1\) in terms of \(x_2\)\[y_1=x_2-\frac{15}{7}\]NOTE: I kind of assumed that the constraint function will make P and Q adjacent because it logically makes sense to me. I double check later that it is the case.\[\]Substituting in all of this into \(A=y_1^2+y_2^2\), we get\[A=\frac{25}{16}x_2^2-\frac{555}{56}x_2+\frac{14625}{784}\]Now, we find the minimum by finding the vertex or by using calc.\[0=\frac{25}{8}x_2-\frac{555}{56}\]\[x_2=\frac{111}{35}\]We know this is the minimum value based on the fact that it's a parabola. Plugging everything back in, we get \[(x_1,y_1)=\left(\frac{27}{35},\frac{36}{35}\right)\]\[(x_2,y_2)=\left(\frac{111}{35},\frac{48}{35}\right)\]And because \(x_2-x_1=y_2+y_1\), we have confirmed that P and Q are adjacent to each other. Plugging in \(x_2\) back into \(A\), we get \(A=\frac{144}{49}\)\[\]\[\]That was my solution. I probably did it the stupid hard way; there's probably an easier Geometry solution.

  13. imqwerty
    • one year ago
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    @dan815

  14. dan815
    • one year ago
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    ah this one is a special triangle so

  15. dan815
    • one year ago
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    |dw:1437071838075:dw|

  16. imqwerty
    • one year ago
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    yes we need to find the min. possible area of the 2 adjacent squares

  17. dan815
    • one year ago
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    |dw:1437072585380:dw|