Solving a Challenging Math Problem | Expert Tips & Strategies

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SUMMARY

This discussion focuses on solving a complex mathematical problem involving the relationship between a rectangle and a reduced ellipse. The original rectangle has dimensions width "w" and height "h", while the reduced ellipse is defined by semi-axes pw/2 and ph/2. The equation governing the ellipse is \(\frac{x^2}{\left(\frac{pw}{2}\right)^2}+ \frac{y^2}{\left(\frac{ph}{2}\right)^2}= 1\). The challenge is to determine the dimensions of a rectangle (Bwa, Bha) that fits within this ellipse, requiring the condition \(\frac{Bwa^2}{p^2w^2}+ \frac{Bha^2}{p^2h^2}= 1\) to be satisfied.

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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
 

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