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Problem 156

# Find the values of $$(x, y, z)$$ that minimize $$F(x, y, z)=x y+2 y z+2 x z$$ given the condition $$\mathrm{G}(\mathrm{x}, \mathrm{y}, z)=\mathrm{xyz}=32$$.

Expert verified
The values of $$(x, y, z)$$ that minimize the function $$F(x, y, z)=xy+2yz+2xz$$ given the constraint $$G(x, y, z)=xyz=32$$ are $$(x, y, z) = (4, 4, 2)$$.
See the step by step solution

## Step 1: Set up the Lagrange's equations

The method of Lagrange multipliers states that we can find the minimum of a function $$F(x,y,z)$$ subject to a constraint $$G(x,y,z)=c$$ by solving the following system of equations: $$\frac{\partial F}{\partial x} = \lambda \frac{\partial G}{\partial x}, \frac{\partial F}{\partial y} = \lambda \frac{\partial G}{\partial y}, \frac{\partial F}{\partial z} = \lambda \frac{\partial G}{\partial z},$$ along with the constraint $$G(x, y, z) = 32$$. In our case: $$F(x, y, z) = xy + 2yz + 2xz$$ $$G(x, y, z) = xyz$$ Now, we calculate the partial derivatives of $$F$$ and $$G$$ with respect to all variables. With respect to $$x$$: $$\frac{\partial F}{\partial x} = y + 2z$$ $$\frac{\partial G}{\partial x} = yz$$ With respect to $$y$$: $$\frac{\partial F}{\partial y} = x + 2z$$ $$\frac{\partial G}{\partial y} = xz$$ With respect to $$z$$: $$\frac{\partial F}{\partial z} = 2y + 2x$$ $$\frac{\partial G}{\partial z} = xy$$

## Step 2: Solve the system of equations

Now, we have the following set of equations: 1. $$y+2z=\lambda yz$$ 2. $$x+2z=\lambda xz$$ 3. $$2y+2x=\lambda xy$$ 4. $$xyz=32$$ We can attempt to solve this system of equations to find the values of $$x, y, z$$, and $$\lambda$$. Divide equation 1 by equation 2: $\frac{y+2z}{x+2z}=\frac{\lambda yz}{\lambda xz} \Rightarrow \frac{y}{x}=\frac{z}{z} \Rightarrow y=x$ Now, rewrite equation 3 using this new information: $2x+2x=\lambda x^2 \Rightarrow 4x=\lambda x^2$ Since $$x\neq 0$$, we can divide both sides by $$x$$: $4=\lambda x$ Now, substitute $$y=x$$ in equation 4: $x^2z=32 \Rightarrow z=\frac{32}{x^2}$ Plug the value of $$z$$ into the equation for $$\lambda$$: $\lambda=\frac{4}{x}$ Now, substitute $$\lambda$$ back into equation 1: $x+2\frac{32}{x^2}=\frac{4}{x}\cdot x\frac{32}{x^2}$ $x+2\frac{32}{x^2}=\frac{128}{x^2}$ $x^3+64=128$ $x^3=64$ $x=4$ Now, using the initial substitution $$y=x$$, we can find the value of $$y$$: $y=4$ Finally, substitute the value of $$x$$ into the equation for $$z$$: $z=\frac{32}{4^2}=\frac{32}{16}=2$

## Step 3: Verify the solution

We have found that the values of $$x, y, z$$ that minimize the function are $$x=4$$, $$y=4$$, and $$z=2$$. To verify the solution, we will check if these values satisfy the constraint: $$G(x, y, z)=xyz=(4)(4)(2)=32$$ The constraint is satisfied, so the values of $$(x, y, z) = (4, 4, 2)$$ indeed minimize the function $$F(x, y, z)$$.

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