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Smells Trigonometric (Posted on 2023-11-12) Difficulty: 2 of 5
x and y are real numbers such that, (4x3-3x)2+(4y3-3y)2=1.

Find the maximum of x+y.

No Solution Yet Submitted by Danish Ahmed Khan    
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Solution Mostly analytical with some help | Comment 1 of 4
... with a little help from Desmos and Wolfram Alpha

Desmos graphical approximation:   appears to be about 1.932

The 4 and the 3 reminded me that there is a trig identity that looks like that.  The title helped as well.  But I had to look it up:

cos(3t) = 4*(cos(t))^3 - 3*cos(t)

Let x = cos(a) and y = cos(b)
We need to maximize cos(a) + cos(b), so it cannot be greater than 2.

(cos(3a))^2 + (cos(3b))^2 = 1

let u = 3a and v = 3b
cos(u)^2 + cos(v)^2 = 1
cos(u)^2 = 1 - cos(v)^2 = sin(v)^2
cos(u) = ± sin(v)
if plus:  v = pi/2 - u
if minus: v = u - pi/2

Take the first one:  
x + y = cos(a) + cos(b) 
= cos(u/3) + cos(v/3)
= cos(u/3) + cos((pi/2 - u)/3)
= cos(u/3) + cos(pi/6 - u/3)
= cos(u/3) + cos(pi/6)cos(u/3) + sin(pi/6)sin(u/3)
f(u) = cos(u/3) + (√3/2)cos(u/3) + (1/2)sin(u/3)
take the derivative 
f'(u) = -(1/3)sin(u/3) - (√3/6)sin(u/3) + (1/6)cos(u/3) = 0

At this point, I let Wolfram do the heavy lifting.
Wolfram Alpha has the maximum as:
√(2 + √3)  or approx 1.93185165258

  Posted by Larry on 2023-11-12 13:33:40
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