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imqwerty
 one year ago
Hi ppl
imqwerty
 one year ago
Hi ppl

This Question is Closed

Study_together
 one year ago
Best ResponseYou've already chosen the best response.0What's your question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hi ^^ have a question ? :)

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2@ganeshie8 @Study_together @Moe95

Jadeishere
 one year ago
Best ResponseYou've already chosen the best response.0I think you mean *People

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes u r correct @Jadeishere

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do P and Q have to be adjacent?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes P and Q have to be adjacent

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let 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 xaxis must correspond to 5 since it's a right triangle. So, I set up a constraint function. \[x_1+y_1+y_2+(5x_2)=5\]\[\implies28x_121x_2=45\]\[\implies x_1=\frac{21x_245}{28}\]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 \(x_2x_1=y_2+y_1\), 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ah this one is a special triangle so

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2yes we need to find the min. possible area of the 2 adjacent squares