Hi ppl

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I think you mean *People
yes u r correct @Jadeishere
Do P and Q have to be adjacent?
yes P and Q have to be adjacent
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.
ah this one is a special triangle so
|dw:1437071838075:dw|
yes we need to find the min. possible area of the 2 adjacent squares
|dw:1437072585380:dw|