Suppose α,β,γ∈[0,π2]\alpha,\beta,\gamma\in [0,\frac{\pi}{2}]α,β,γ∈[0,2π] and cosα+cosβ+cosγ=1\cos\alpha +\cos\beta +\cos\gamma=1cosα+cosβ+cosγ=1.
How to get the maximum and minimum of the expression sin4α+sin4β+sin4γ\sin^{4}\alpha+\sin^{4}\beta+\sin^{4}\gammasin4α+sin4β+sin4γ ?
Any good ideas?
Note by Haosen Chen 3 years ago
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
_italics_
**bold**
__bold__
- bulleted- list
1. numbered2. list
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"
\(
\)
\[
\]
2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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}(1−x2)2+(1−y2)2+(1−z2)2 given that x,y,z∈[0,1]x,y,z\in[0,1]x,y,z∈[0,1] and x+y+z=1x+y+z=1x+y+z=1.
Let x+y=tx+y=tx+y=t where ttt 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−y2)2+(1−z2)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}(1−x2)2+(1−(t−x)2)2+(1−(1−t)2)2,which can be seen as a function of xxx,say, f(x)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)f′(x)=4(2x−t)(x2−tx+t2−1)=8=8=8(x−t−4−3t22)(x-\frac{ t-\sqrt{4-3t^{2}} }{2})(x−2t−4−3t2)(x−t2)(x-\frac{t}{2})(x−2t)(x−t+4−3t22)(x-\frac{ t+\sqrt{4-3t^{2}} }{2})(x−2t+4−3t2).
Since 0≤t≤10\le t \le 10≤t≤1,it’s clear thatt−4−3t22≤0\frac{ t-\sqrt{4-3t^{2}} }{2}\le 02t−4−3t2≤0 and 1≤t+4−3t221\le \frac{ t+\sqrt{4-3t^{2}} }{2}1≤2t+4−3t2.
Thus,f(x)f(x)f(x) is increasing on the interval (0,t2)(0,\frac{t}{2})(0,2t) and decreasing on the interval (t2,t)(\frac{t}{2},t)(2t,t).
Hence, every fixed parameter ttt,the maximum and minimum of f(x)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}f(2t)=2(1−(2t)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}f(0)(=f(t))=1+(1−t2)2+(1−(1−t)2)2 respectively,which can also be seen as functions of ttt,say,g(t)g(t)g(t) and h(x)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)g′(t)=29t(t−32)(t−2),h′(t)=8t(t−12)(t−1)h'(t)=8t(t-\frac{1}{2})(t-1)h′(t)=8t(t−21)(t−1).
So we see that g(t)≤g(23)=6427g(t)\le g(\frac{2}{3}) = \frac{64}{27}g(t)≤g(32)=2764 and h(t)≥h(0)=2h(t)\ge h(0) =2h(t)≥h(0)=2,for all 0≤t≤10\le t \le 10≤t≤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}(1−x2)2+(1−y2)2+(1−z2)2 are 6427\frac{64}{27}2764 and 222,respectively.
How about putting cosα=a,cosβ=b,cosγ=c\cos\alpha=a,\cos\beta=b,\cos\gamma=ccosα=a,cosβ=b,cosγ=c,so a+b+c=1a+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(1−a2)2+(1−b2)2+(1−c2)2
Log in to reply
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.
Problem Loading...
Note Loading...
Set Loading...
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
The problem is equivalent to determine the maximum and minimum of (1−x2)2+(1−y2)2+(1−z2)2 given that x,y,z∈[0,1] and x+y+z=1.
Let x+y=t where t is a fixed parameter. Then (1−x2)2+(1−y2)2+(1−z2)2=(1−x2)2+(1−(t−x)2)2+(1−(1−t)2)2,which can be seen as a function of x,say, f(x).
Its derivative is f′(x)=4(2x−t)(x2−tx+t2−1)=8(x−2t−4−3t2)(x−2t)(x−2t+4−3t2).
Since 0≤t≤1,it’s clear that2t−4−3t2≤0 and 1≤2t+4−3t2.
Thus,f(x) is increasing on the interval (0,2t) and decreasing on the interval (2t,t).
Hence, every fixed parameter t,the maximum and minimum of f(x) are f(2t)=2(1−(2t)2)2+(1−(1−t)2)2 and f(0)(=f(t))=1+(1−t2)2+(1−(1−t)2)2 respectively,which can also be seen as functions of t,say,g(t) and h(x) respectively. Again,culculate their derivatves: g′(t)=29t(t−32)(t−2),h′(t)=8t(t−21)(t−1).
So we see that g(t)≤g(32)=2764 and h(t)≥h(0)=2,for all 0≤t≤1.
Therefore, the maximum and minimum of (1−x2)2+(1−y2)2+(1−z2)2 are 2764 and 2,respectively.
How about putting cosα=a,cosβ=b,cosγ=c,so a+b+c=1 and then find the max and min of (1−a2)2+(1−b2)2+(1−c2)2
Log in to reply
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.