Square Inscribed in a Square: Maximizing Distance Between Vertices

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To determine the greatest distance between a vertex of a square inscribed in another square, one must consider the dimensions of both squares. The inner square has a perimeter of 20, giving it a side length of 5, while the outer square, with a perimeter of 28, has a side length of 7. The maximum distance between a vertex of the inner square and a vertex of the outer square occurs when the inner square is positioned diagonally within the outer square. The discussion clarifies that an inscribed square typically touches the sides of the outer square at its vertices, rather than along the base. Understanding these geometric relationships is crucial for solving the problem effectively.
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Homework Statement



A square of perimeter 20 is inscribed in a square of perimeter 28. What is the greatest distance between a vertex of the inner square and a vertex of the outer square.

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The Attempt at a Solution

I have a question. Can a square be inscribed in another square by having it sit along the base and the side of the bigger square? Does it only have to touch vertice to side?
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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