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anygoal [31]
3 years ago
12

Use the method of Lagrange Multipliers to End the maximum and minimum values of the function subject to the given constraints: f

(x, y, z) = yz + xy xy = 1 y^2 + z^2 = 1
Mathematics
1 answer:
Alex Ar [27]3 years ago
6 0

Looks like you're looking for the extrema of f(x,y,z)=yz+xy subject to xy=1 and y^2+z^2=1. The Lagrangian is

L(x,y,z,\lambda,\mu)=yz+xy+\lambda(xy-1)+\mu(y^2+z^2-1)

Look for any critical points:

L_x=y+\lambda y=y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1

L_y=z+x+\lambda x+2\mu y=0

L_z=y+2\mu z=0

L_\lambda=xy-1=0\implies xy=1

L_\mu=y^2+z^2-1=0\implies y^2+z^2=1

  • If y=0, then

L_z=0\implies2\mu z=0\implies z=0 (we don't want \lambda=\mu=0)

but then y^2+z^2=0\neq1 so we omit this case.

  • If \lambda=-1, then

L_y=0\implies z+2\mu y=0

L_\mu=0\implies(1+4\mu^2)y^2-1=0\implies y=\pm\dfrac1{\sqrt{1+4\mu^2}}

L_z=0\implies z=\mp\dfrac1{2\mu\sqrt{1+4\mu^2}}

L_\mu=0\implies\dfrac1{1+4\mu^2}+\dfrac1{4\mu^2(1+4\mu^2)}=1\implies\mu=\pm\dfrac12

L_\lambda=0\implies x=\pm\sqrt{1+4\mu^2}

Now,

  • if \mu=\dfrac12, we have two critical points at \left(\sqrt2,\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right) and \left(-\sqrt2,-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right);
  • if \mu=-\dfrac12, we have two additional critical points \left(\sqrt2,\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right) and \left(-\sqrt2,-\dfrac1{\sqrt2},-\dfrac1{\sqrt2}\right)

The first two critical points give a minimum value of f(x,y,z)=\dfrac12, and the other two give a maximum value of f(x,y,z)=\dfrac32.

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Answer:

This question seems to be asking for the work, so I put it below.

Anyhow, Marla had 3.5 candies, and Ron had 10.5.

Step-by-step explanation:

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