MHB Solving a Challenging Math Problem | Expert Tips & Strategies

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The discussion focuses on solving a complex math problem involving a rectangle inscribed within a reduced ellipse. The original rectangle has dimensions defined by width "w" and height "h," while the ellipse is characterized by semi-axes of length "pw/2" and "ph/2." A coordinate system is established with the ellipse's center as the origin, and the upper left corner of the rectangle is positioned at (Bwa/2, Bha/2). The key equation to solve is \frac{Bwa^2}{p^2w^2}+ \frac{Bha^2}{p^2h^2}= 1, which ensures that the rectangle fits within the ellipse. Participants are encouraged to ask questions for further clarification and assistance.
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Take the original rectangle to have width "w" and height "h". Then the "reduced" ellipse has axe of length pw and ph, semi-axes pw/2 and ph/2, and is given by the equation \frac{x^2}{\left(\frac{pw}{2}\right)^2}+ \frac{y^2}{\left(\frac{ph}{2}\right)^2}= \frac{4x^2}{p^2w^2}+ \frac{4y^2}{p^2h^2}= 1.

Now, we have a rectangle Bwa by Bha that fits inside that reduced ellipse. Setting up a coordinate system with origin at the center of the ellipse, coordinate axes along the axes of the ellipse. In that coordinate system, the upper left corner of the rectangle has coordinates (Bwa/2, Bha/2) and, since that corner lies on the ellipse, must satisfy the equation of the ellipse. That is, we must have \frac{Bwa^2}{p^2w^2}+ \frac{Bha^2}{p^2h^2}= 1
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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