Use Lagrange multipliers to find the points on the given surface $y^{2}=16+xz$ that are closest to the origin. $\newline$ {:[ $(\text{smaller } y\text{-value}) \quad(x,y,z)=(\square)$ ],[ $(\text{larger } y\text{-value}) \quad(x,y,z)=(\square)$ ]:} $\newline$ Submit Answer

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Find Coordinate on Unit Circle

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Q. Use Lagrange multipliers to find the points on the given surface

y^{2}=16+xz

that are closest to the origin.

\newline

{:[

(\text{smaller } y\text{-value}) \quad(x,y,z)=(\square)

],[

(\text{larger } y\text{-value}) \quad(x,y,z)=(\square)

]:}

\newline

Submit Answer

Set up function and constraint: Set up the function to minimize and the constraint equation. $\newline$ We want to minimize the distance to the origin, which is given by the function $f(x,y,z) = x^2 + y^2 + z^2$ . The constraint is given by the equation $g(x,y,z) = y^2 - (16 + xz) = 0$ .
Use Lagrange multipliers: Set up the system of equations using Lagrange multipliers. $\newline$ We introduce a Lagrange multiplier $\lambda$ and set up the following system of equations: $\newline$ $\newlineabla f(x,y,z) = \lambda $\newline$ abla g(x,y,z)$ $\newline$ This gives us the following equations: $\newline$ $2x = \lambda(-z)$ ( $1$ ) $\newline$ $2y = \lambda(2y)$ ( $2$ ) $\newline$ $2z = \lambda(-x)$ ( $3$ ) $\newline$ And the constraint equation: $\newline$ $y^2 = 16 + xz$ ( $4$ )
Solve system of equations: Solve the system of equations. $\newline$ From equation ( $1$ ), we have $x = -\lambda z/2$ . From equation ( $3$ ), we have $z = -\lambda x/2$ . Substituting $z$ from equation ( $3$ ) into equation ( $1$ ), we get $x = -\lambda(-\lambda x/4)/2$ , which simplifies to $x(\lambda^2 + 4) = 0$ . This gives us two cases: $x = 0$ or $\lambda = \pm 2i$ , where $i$ is the imaginary unit. Since we are looking for real solutions, we discard $\lambda = \pm 2i$ and consider $x = 0$ .
Substitute $x=0$ into constraint: Substitute $x = 0$ into the constraint equation. $\newline$ Substituting $x = 0$ into equation ( $4$ ), we get $y^2 = 16$ . This gives us two possible values for $y$ : $y = \pm4$ .
Find corresponding z values: Find the corresponding z values. Since $x = 0$ , the constraint equation becomes $y^2 = 16 + 0\cdot z$ , which we have already solved for $y$ . Now we need to find $z$ when $y = \pm 4$ . From equation ( $3$ ), if $x = 0$ , then $2z = 0$ , which means $z = 0$ .
Compile the solutions: Compile the solutions. $\newline$ We have found two points that satisfy the constraint and could potentially minimize the distance to the origin: $\newline$ (smaller y-value) $(x,y,z) = (0, -4, 0)$ $\newline$ (larger y-value) $(x,y,z) = (0, 4, 0)$