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cours/Excalidraw/problème de la plus grande diagonale d'une valise aux bords arrondis 2026-04-16.excalidraw.md
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==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'

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le tableau de variations est évident étant donné le problème considéré. On pourra toujours montrer que cet unique extrémum est bien un maximum à partir du fait que 𝑎 ∈ [0; π/2] ^OnHRRJcd

Embedded Files

8923efd922b888d467179262d7b788641d043cc7: l

b3d5701da7f512bb6c2cfeefa687ef9448961a27: L

a745f1dc9892805697eb7dc6df5134ce60d3e778: \alpha

b76ee720d7e0bdf63f408ee39db94f3d31ca9f0e: r

13099823a77cd5a2252736e0f029c959c6ee8e2c: =\begin{pmatrix}l\\L\end{pmatrix}

e8b51e7e123fde3009046ff7524c05292519716e: =\begin{pmatrix}r \cos \alpha\\ r \sin \alpha\end{pmatrix}

2eb73fd7a8bd07b84080b3bb149f4819930587be: = \begin{pmatrix}l + r \cos \alpha\\ L + r \sin \alpha\end{pmatrix}

4dd7657e8aff3434304e0c34035d8f6e3e9c7c5d: \text{donc :}

ce7a3ec53f89ae18583ec0042233662e411ba23b: \begin{align} \left\| \begin{pmatrix}l+r\cos \alpha\\L+r \sin \alpha\end{pmatrix} \right\| &= l^{2}+2lr\cos\alpha + r^{2}\cos ^{2}\alpha + L^{2} + 2Lr\sin\alpha + r^{2}\sin ^{2}\alpha \\&= (l^{2}+L^{2}+r^{2}) + 2r(l\cos\alpha+L\sin\alpha)\end{align}

16647f064fb7a498cb3e96cfa483672e387b3557: \text{on veut donc maximiser } f: f(\alpha) = l\cos\alpha+L\sin\alpha

c07e13d38824e5b0021dd5fdfadb2c55ad79c69a: f'(\alpha) = -l\sin\alpha + L\cos\alpha

fafa660ca9b7e22e4de89da24218f9436103d840: \begin{align} f'(\alpha) = 0 &\iff L \cos \alpha = l \sin \alpha \\&\iff \frac{L}{l} \frac{\cos \alpha}{\sin\alpha} = 1 \\&\iff \tan \alpha = \frac{l}{L} \\&\iff \boxed {\alpha = \arctan \frac{l}{L}} \end{align}

%%

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