Find Maximum of xyz: Real Solutions

[solakis1]

Find the maximum of xyz when $y^2=xz$ and $ x+y+z=a$

x,y,z are reals

 

[jvicens]

Hello everyone,

I have been working on a problem and I am stuck. I am trying to find the maximum value of the product xyz when y^2=xz and x+y+z=a, where x,y,z are real numbers. I have tried using derivatives and setting up equations, but I am not getting anywhere. Can someone please help me out?

I would approach this problem by first understanding the given constraints and then finding a way to optimize the given equation.

Firstly, we know that y^2=xz and x+y+z=a. This means that we can substitute y^2 for xz in the second equation, giving us y^2+y+z=a. We can then rearrange this equation to get z=a-y-y^2.

Next, we can substitute this value of z into the first equation, giving us y^2=x(a-y-y^2). Simplifying this, we get y^2+xy+y^3=xa.

To find the maximum value of xyz, we can use the AM-GM inequality, which states that the arithmetic mean of a set of numbers is always greater than or equal to the geometric mean of the same set of numbers. In this case, the numbers are y, x, and y^3. Using this inequality, we can write:

(x+y+y^3)/3 ≥ (xy^3)^(1/3)

Simplifying this, we get:

x+y+y^3 ≥ 3xy^(1/3)

Now, using the fact that y^2=xz, we can substitute xz for y^2 in the above equation, giving us:

x+z+y^3 ≥ 3xz^(1/3)

Finally, substituting z=a-y-y^2 and simplifying, we get:

x+a-y-y^2+y^3 ≥ 3(a-y-y^2)^(1/3)

To find the maximum value of xyz, we need to find the maximum value of this expression. To do this, we can use calculus and take the derivative of this expression with respect to y. Setting the derivative equal to 0 and solving for y, we get y=1.

Substituting y=1 back into our original equation, we get x+z=1. Since we want to find the maximum value of xyz, we can set z=0 and x=1, giving us the maximum value of xyz as 0