Difficult max and min

Suppose \(\alpha,\beta,\gamma\in [0,\frac{\pi}{2}]\) and \(\cos\alpha +\cos\beta +\cos\gamma=1\).

How to get the maximum and minimum of the expression sin⁡4α+sin⁡4β+sin⁡4γ\sin^{4}\alpha+\sin^{4}\beta+\sin^{4}\gamma ?

Any good ideas?

#Algebra

Note by Haosen Chen
3 years ago

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Comments

The problem is equivalent to determine the maximum and minimum of (1−x2)2+(1−y2)2+(1−z2)2(1-x^{2})^{2}+(1-y^{2})^{2}+(1-z^{2})^{2} given that x,y,z∈[0,1]x,y,z\in[0,1] and x+y+z=1x+y+z=1.

Let x+y=tx+y=t where tt is a fixed parameter. Then (1−x2)2+(1−y2)2+(1−z2)2(1-x^{2})^{2}+(1-y^{2})^{2}+(1-z^{2})^{2}=(1−x2)2+(1−(t−x)2)2+(1−(1−t)2)2(1-x^{2})^{2}+(1-(t-x)^{2})^{2}+(1-(1-t)^{2})^{2},which can be seen as a function of xx,say, f(x)f(x).

Its derivative is f′(x)=4(2x−t)(x2−tx+t2−1)f'(x)=4(2x-t)(x^{2}-tx+t^{2}-1)=8=8(x−t−4−3t22)(x-\frac{ t-\sqrt{4-3t^{2}} }{2})(x−t2)(x-\frac{t}{2})(x−t+4−3t22)(x-\frac{ t+\sqrt{4-3t^{2}} }{2}).

Since 0≤t≤10\le t \le 1,it’s clear thatt−4−3t22≤0\frac{ t-\sqrt{4-3t^{2}} }{2}\le 0 and 1≤t+4−3t221\le \frac{ t+\sqrt{4-3t^{2}} }{2}.

Thus,f(x)f(x) is increasing on the interval (0,t2)(0,\frac{t}{2}) and decreasing on the interval (t2,t)(\frac{t}{2},t).

Hence, every fixed parameter tt,the maximum and minimum of f(x)f(x) are f(t2)=2(1−(t2)2)2+(1−(1−t)2)2f(\frac{t}{2})=2(1-(\frac{t}{2})^{2})^{2}+(1-(1-t)^{2})^{2} and f(0)(=f(t))=1+(1−t2)2+(1−(1−t)2)2f(0)(=f(t))=1+(1-t^{2})^{2}+(1-(1-t)^{2})^{2} respectively,which can also be seen as functions of tt,say,g(t)g(t) and h(x)h(x) respectively. Again,culculate their derivatves: g′(t)=92t(t−23)(t−2)g'(t)=\frac{9}{2}t(t-\frac{2}{3})(t-2),h′(t)=8t(t−12)(t−1)h'(t)=8t(t-\frac{1}{2})(t-1).

So we see that g(t)≤g(23)=6427g(t)\le g(\frac{2}{3}) = \frac{64}{27} and h(t)≥h(0)=2h(t)\ge h(0) =2,for all 0≤t≤10\le t \le 1.

Therefore, the maximum and minimum of (1−x2)2+(1−y2)2+(1−z2)2(1-x^{2})^{2}+(1-y^{2})^{2}+(1-z^{2})^{2} are 6427\frac{64}{27} and 22,respectively.

Haosen Chen - 2 years, 12 months ago

How about putting cos⁡α=a,cos⁡β=b,cos⁡γ=c\cos\alpha=a,\cos\beta=b,\cos\gamma=c,so a+b+c=1a+b+c=1 and then find the max and min of (1−a2)2+(1−b2)2+(1−c2)2(1-a^2)^2+(1-b^2)^2+(1-c^2)^2

X X - 3 years ago

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Oh,thanks for attention,it does work.In fact,I have solved this problem already(on yesterday's morning) and I will put my solution here this Sunday.

Haosen Chen - 3 years ago
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