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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.

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LammettHash
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.

Using Lagrange multipliers, it is found that the numbers are: x = y = z = 40.

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The product is that we want to maximize is:

[tex]f(x,y,z) = xyz[/tex]

The constraint is:

[tex]x + y + z = 120[/tex]

[tex]g(x,y,z) = x + y + z - 120[/tex]

The gradients are:

[tex]\nabla_f = (yz, xz, xy)[/tex]

[tex]\nabla_g = (1,1,1)[/tex]

Their relation is:

[tex]\nable_f = \lambda\nabla_g[/tex]

[tex](yz, xz, xy) = \lambda(1,1,1)[/tex]

Then, the system is:

[tex]yz = \lambda[/tex]

[tex]xz = \lambda[/tex]

[tex]xy = \lambda[/tex]

And thus, x = y = z.

[tex]x + y + z = 120[/tex]

Thus:

[tex]3x = 120[/tex]

[tex]x = \frac{120}{3}[/tex]

[tex]x = 40[/tex]

Thus, the values are x = y = z = 40.

A similar problem is given at https://brainly.com/question/4609414

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