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Gnesinka [82]
3 years ago
15

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.
Mathematics
1 answer:
murzikaleks [220]3 years ago
7 0
We're maximizing xyz subject to x+y+z=120. We have Lagrangian

L(x,y,z,\lambda)=xyz+\lambda(x+y+z-120)

with partial derivatives (set to 0)

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

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

Since x,y,z>0, we arrive at x=y=z, which means

x+y+z=3x=120\implies x=y=z=40

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