Find the minimum of M M

Algebra Level 5

If { a , b , c , d > 0 c 2 + d 2 = ( a 2 + b 2 ) 3 a 3 c + b 3 d = M \begin{cases}a,b,c,d>0\\c^2+d^2=(a^2+b^2)^3\\\dfrac{a^3}{c}+\dfrac{b^3}{d}=M\end{cases} find the minimum value of M M correct to 3 3 decimal places.


The answer is 1.000.

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1 solution

Here is my solution.

Let us introduce u , v u,v such that cos = a / a 2 + b 2 , cos v = c / c 2 + d 2 \cos = a/\sqrt{a^2+b^2},\ \cos v = c/\sqrt{c^2+d^2} . Clearly, the constraints a , b , c , d > 0 a,b,c,d>0 impose the constraints u , v [ 0 , π / 2 ] u,v\in [0,\pi/2] . Note that now one can write M = a 3 c + b 3 d = cos 3 u cos v + sin 3 u sin v , M = \frac{a^3}{c}+\frac{b^3}{d}=\frac{\cos^3u}{\cos v}+\frac{\sin^3 u}{\sin v}, which follows from the given constraint c 2 + d 2 = ( a 2 + b 2 ) 3 c^2+d^2=(a^2+b^2)^3 . Therefore, minimizing M M is equivalent to minimizing the above expression in u , v u,v which is an unconstrained optimization problem except for the trivial constraint u , v [ 0 , π / 2 ] u,v\in [0,\pi/2] . The minima can be easily calculated to be the point where u = v u=v , where the value of M M becomes cos 2 u + sin 2 u = 1 \cos^2u+\sin^2u=1 .

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