Suppose you have 5 real numbers whose sum

of squares is equal to 5. What is the maximum
value of the sum of cubes of these 5 numbers?

Please show step

You don't say what kind of numbers they are ,

e.g. if they are rationals
a=1
b = .5
c= 1.1
d= 1.4
sum of squares of those 4 = 4.42
then the 5th number is √.58

if they are integers, they must be
±1,±1,±1,±1,±1 or ±2,±1,0,0,0

e.g. 2^2 + 1^2 + 0^2 + 0^2 = 5

The sum of cubes of the first set is 5
the sum of cubes of the second set is 9

so the numbers must be 2, 1, 0, 0, 0

incidentally the sum of cubes of my first example would be
5.6417...

On the other hand, suppose the 5 numbers are
√5, 0, 0, 0, 0
the sum of their squares is indeed 5
and the sum of their cubes is (√5)^3 or 11
18...

how about that?

Was there more information?

and the sum of their cubes is (√5)^3 or 11.18...

To find the maximum value of the sum of cubes of 5 real numbers, we can use the method of Lagrange multipliers.

Let's denote the 5 real numbers as x1, x2, x3, x4, and x5. We need to find the maximum value of the sum of cubes, which is given by the function F(x1, x2, x3, x4, x5) = x1^3 + x2^3 + x3^3 + x4^3 + x5^3.

However, we have the constraint that the sum of squares of these 5 numbers is equal to 5, which can be expressed as g(x1, x2, x3, x4, x5) = x1^2 + x2^2 + x3^2 + x4^2 + x5^2 - 5 = 0.

To apply Lagrange multipliers, we introduce a Lagrange multiplier λ, and form the new function H(x1, x2, x3, x4, x5, λ) = F(x1, x2, x3, x4, x5) + λg(x1, x2, x3, x4, x5).

Taking the partial derivatives of H with respect to each variable and equating them to 0 will give us the critical points of H.

∂H/∂x1 = 3x1^2 + 2λx1 = 0
∂H/∂x2 = 3x2^2 + 2λx2 = 0
∂H/∂x3 = 3x3^2 + 2λx3 = 0
∂H/∂x4 = 3x4^2 + 2λx4 = 0
∂H/∂x5 = 3x5^2 + 2λx5 = 0
∂H/∂λ = x1^2 + x2^2 + x3^2 + x4^2 + x5^2 - 5 = 0

From the above equations, we can see that each x1, x2, x3, x4, x5, and λ satisfy their respective equations, which imply:

x1^2 + 2λx1 = 0
x2^2 + 2λx2 = 0
x3^2 + 2λx3 = 0
x4^2 + 2λx4 = 0
x5^2 + 2λx5 = 0
x1^2 + x2^2 + x3^2 + x4^2 + x5^2 - 5 = 0

Since the sum of the squares of the numbers is equal to 5, we can solve the last equation to find λ as follows:

x1^2 + x2^2 + x3^2 + x4^2 + x5^2 - 5 = 0
x1^2 + x2^2 + x3^2 + x4^2 + (5 - x1^2 - x2^2 - x3^2 - x4^2) - 5 = 0
2(x1^2 + x2^2 + x3^2 + x4^2) - 10 = 0
x1^2 + x2^2 + x3^2 + x4^2 = 5/2

Substituting this value of x1^2 + x2^2 + x3^2 + x4^2 into any of the equations involving λ, we can solve for λ.

For simplicity, let's assume x1^2 + x2^2 + x3^2 + x4^2 = 5/2. Then:

x1^2 + 2λx1 = 0
x2^2 + 2λx2 = 0
x3^2 + 2λx3 = 0
x4^2 + 2λx4 = 0

Since the maximum value of x1^2 + x2^2 + x3^2 + x4^2 is 5/2, the maximum value of λ can be found by substituting 5/2 for x1^2 + x2^2 + x3^2 + x4^2 in any of the above equations.

Once you find the value of λ, you can solve the remaining equations simultaneously to find the values of x1, x2, x3, and x4 that satisfy these equations.

Finally, the maximum value of the sum of cubes, F(x1, x2, x3, x4, x5), can be found by substituting the values of x1, x2, x3, x4, and x5 into the function F(x1, x2, x3, x4, x5) = x1^3 + x2^3 + x3^3 + x4^3 + x5^3.