SOLUTION: Let x, y, and z be real numbers. If x^2 + y^2 + z^2 = 1, then find the maximum value of 3x + 4y + 5z + x^3 + \frac{4x^2*y)/{z} + \frac{z^5}{xy^2}

Algebra.Com
Question 1209923: Let x, y, and z be real numbers. If x^2 + y^2 + z^2 = 1, then find the maximum value of
3x + 4y + 5z + x^3 + \frac{4x^2*y)/{z} + \frac{z^5}{xy^2}

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let f(x,y,z)=3x+4y+5z+x
3
+
z
4x
2
y

+
xy
2

z
5


and g(x,y,z)=x
2
+y
2
+z
2
−1=0.
We use the method of Lagrange multipliers to find the maximum value of f(x,y,z) subject to the constraint g(x,y,z)=0.
We need to solve the system of equations:
∇f=λ∇g
g(x,y,z)=0
First, let's find the gradients:
∇f=(3+3x
2
+
z
8xy


x
2
y
2

z
5


,4+
z
4x
2



xy
3

2z
5


,5−
z
2

4x
2
y

+
xy
2

5z
4


)
∇g=(2x,2y,2z)
Now, we set ∇f=λ∇g:
3+3x
2
+
z
8xy


x
2
y
2

z
5


=2λx
4+
z
4x
2



xy
3

2z
5


=2λy
5−
z
2

4x
2
y

+
xy
2

5z
4


=2λz
x
2
+y
2
+z
2
=1
This system of equations is highly nonlinear and difficult to solve analytically. However, we can make some observations.
If we consider the case where x,y,z>0, we have:
When x=0.6, y=0.8, and z is a small positive number, the term
z
4x
2
y

becomes very large.
When x,y are small positive numbers and z is close to 1, the term
xy
2

z
5


becomes very large.
This suggests that the maximum value might occur when x,y,z are all positive and have a certain relationship.
That's the only answer I can give for now!

Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
.
Let x, y, and z be real numbers. If x^2 + y^2 + z^2 = 1, then find the maximum value of
3x + 4y + 5z + x^3 + + .
~~~~~~~~~~~~~~~~~~~~~~~~~~~


            @CPhill finds it difficult to give a definitive answer in his post.


Meanwhile,  the answer to this question is very simple:  under given conditions,
the given function/expression has  NO  maximum.

It is because the term    of the expression has variable  z  in the denominator.


        Take  (x,y,z)  in vicinity of  (,,),   so that   x^2 + y^2 + z^2 = 1   is valid,

                  and let  z  goes to zero from the positive side.


Then the term    tends to positive infinity,
and with this term, the whole expression tends to positive infinity.


////////////////////////////////


So the answer in the post by @CPhill is empty.



RELATED QUESTIONS

Find the smallest possible value of $$\frac{(y-x)^2}{(y-z)(z-x)} +... (answered by Alan3354)
(a) Let x, y, and z be positive real numbers. Find the largest possible value of... (answered by Edwin McCravy,ikleyn,mccravyedwin)
If \[\frac{x}{y} = \frac{4}{5}, \; \frac{y}{z} = \frac{2}{5}, \;\text{and} \; \frac{z}{w} (answered by ikleyn,josgarithmetic,greenestamps,math_tutor2020)
Find the maximum p such that 2x^4 y^2 + \frac{9}{4} y^4 z^2 + \frac{3}{4} z^4 x^2 + 3x^3 (answered by mccravyedwin)
Let be three positive numbers such that: x^2 + y^2 + z^2 = 2(xy + xz + yz). x + y... (answered by CPhill,ikleyn)
Solve for the variable $x$ in terms of $y$ and $z$: xy + 2x = \frac{x - 2y + z}{3} + (x... (answered by yurtman)
Find the minimum value of the product P(x,y,z) = (2x + 3y)(x + 4z) \left( y +... (answered by CPhill)
If x+y+z=16 then find the maximum value of (x-3)(y-5)(z-2). Given that (x-3)>0,... (answered by Edwin McCravy,ikleyn)
Let x and y be real numbers satisfying \frac{x^2y^2 - 1}{2y - 1} = 4x + y. Find the... (answered by CPhill,ikleyn,math_tutor2020)