Chapter 6: Chapter 6

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 $\mathrm{G}(\mathrm{x}, \mathrm{y}, z)=\mathrm{xyz}=32$.

Short Answer

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 $(x, y, z) = (4, 4, 2)$.

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TABLE OF CONTENTS :

TABLE OF CONTENTS

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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Chapter 6: Chapter 6

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

Short Answer

Step by step solution

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Step 1: Set up the Lagrange's equations

Step 2: Solve the system of equations

Step 3: Verify the solution

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Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 7

Chapter 8

Chapter 9

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Chapter 13

Chapter 14

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Chapter 18

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