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