Use lagrange multipliers to find three positive numbers whose sum is 120 and whose product is maximum. (enter your answers as a comma-separated list.) (40,40,40) incorrect: your answer is incorrect.

We're maximizing [tex]xyz[/tex] subject to [tex]x+y+z=120[/tex]. We have Lagrangian

[tex]L(x,y,z,\lambda)=xyz+\lambda(x+y+z-120)[/tex]

with partial derivatives (set to 0)

[tex]L_x=yz+\lambda=0\implies yz=-\lambda[/tex]
[tex]L_y=xz+\lambda=0\implies xz=-\lambda[/tex]
[tex]L_z=xy+\lambda=0\implies xy=-\lambda[/tex]
[tex]L_\lambda=x+y+z-120=0[/tex]

[tex]yz=xz\implies z(x-y)=0\implies z=0\text{ or }x=y[/tex]
[tex]xz=xy\implies x(y-z)=0\implies x=0\text{ or }y=z[/tex]
[tex]yz=xy\implies y(x-z)=0\implies y=0\text{ or }x=z[/tex]

Since [tex]x,y,z>0[/tex], we arrive at [tex]x=y=z[/tex], which means

[tex]x+y+z=3x=120\implies x=y=z=40[/tex]

and we get a maximum value of [tex]40^3=64000[/tex] for the product.