Question 495663
If a and b are the side lengths with a+b = 50, we can find the minimum possible value of a diagonal several ways:

If you know power means, we have *[tex \LARGE \sqrt{\frac{a^2+b^2}{2}} \ge \frac{a+b}{2} \Rightarrow \sqrt{\frac{a^2+b^2}{2}} \ge 25 \Rightarrow \sqrt{a^2+b^2} \ge 25\sqrt{2}], equality occurs when a=b (square).
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The Cauchy-Schwarz inequality states that

*[tex \LARGE (a^2+b^2)(1+1) \ge (a+b)^2]

*[tex \LARGE (a^2+b^2)(2) \ge 50^2 \Rightarrow \sqrt{a^2+b^2} \ge 25\sqrt{2}]

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We can say that *[tex \LARGE (a+b)^2 = a^2 + b^2 + 2ab = d^2 + 2ab], the LHS is constant so we can maximize 2ab (twice the area of the rectangle) to minimize the diagonal d. We can show several ways the optimal case is a=b.
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All of these solutions yield *[tex \LARGE 25\sqrt{2}] as the minimal possible diagonal.