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MTH 254 Calculus IV Winter 2026

Richard Zhao

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Worksheet Assigned Problems for Section 14.8

The problems listed below are assigned to be included in your problem set portfolio. Note that a specific selection of these problems will also form the written homework assignments. I recommend working through all of them to ensure a solid grasp of the material. Reach out to Richard for help if you get stuck or have any questions.
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The solutions will be posted after the written homework due dates. If you have any questions about your work, talk to Richard and he is happy to discuss the process with you.
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14.8.1.

Find the extreme values of the function \(f(x,y) = 2x + 4y\) subject to the constraint \(g(x,y) = x^2 + y^2 - 5 = 0\) following the following steps:
  1. Show that the Lagrange equation \(\nabla f = \lambda \nabla g\) gives \(\lambda x = 1\) and \(\lambda y = 2\text{.}\)
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  2. Show that these equations imply \(\lambda \neq 0\) and \(y = 2x\text{.}\)
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  3. Use the constraint equation to determine the possible critical points \((x,y)\text{.}\)
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  4. Evaluate \(f(x,y)\) at the critical points and determine the minimum and maximum values.
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Solution.
  1. We are given the objective function \(f(x,y) = 2x + 4y\) and the constraint \(g(x,y) = x^2 + y^2 - 5 = 0\text{.}\) Taking the gradients, we get:
    \begin{align*} \nabla f \amp = \langle 2, 4 \rangle \\ \nabla g \amp = \langle 2x, 2y \rangle \end{align*}
    Setting \(\nabla f = \lambda \nabla g\) gives us the system:
    \begin{align*} 2 \amp = \lambda (2x) \implies 1 = \lambda x \\ 4 \amp = \lambda (2y) \implies 2 = \lambda y \end{align*}
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  2. If \(\lambda = 0\text{,}\) the first equation becomes \(1 = 0\text{,}\) which is a contradiction. Therefore, \(\lambda \neq 0\text{.}\) Since \(\lambda \neq 0\text{,}\) we can solve for \(x\) and \(y\) in terms of \(\lambda\text{:}\)
    \begin{align*} x \amp = \frac{1}{\lambda} \\ y \amp = \frac{2}{\lambda} \end{align*}
    Notice that \(y = \frac{2}{\lambda} = 2\left(\frac{1}{\lambda}\right) = 2x\text{.}\) Thus, \(y = 2x\text{.}\)
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  3. Substitute \(y = 2x\) into the constraint equation \(x^2 + y^2 = 5\text{:}\)
    \begin{align*} x^2 + (2x)^2 \amp = 5 \\ x^2 + 4x^2 \amp = 5 \\ 5x^2 \amp = 5 \implies x^2 = 1 \implies x = \pm 1 \end{align*}
    When \(x = 1\text{,}\) \(y = 2(1) = 2\text{.}\) When \(x = -1\text{,}\) \(y = 2(-1) = -2\text{.}\) Our critical points are \((1,2)\) and \((-1,-2)\text{.}\)
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  4. Finally, evaluate \(f(x,y) = 2x + 4y\) at the critical points:
    \begin{align*} f(1,2) \amp = 2(1) + 4(2) = 10 \\ f(-1,-2) \amp = 2(-1) + 4(-2) = -10 \end{align*}
    The maximum value is \(10\) at \((1,2)\text{,}\) and the minimum value is \(-10\) at \((-1,-2)\text{.}\)
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Exercise Group.

In the following exercises, find the minimum and maximum values of the function subject to the given constraints.
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14.8.7.
\(f(x,y) = xy \, , \qquad 4x^2 + 9y^2 = 32\)
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Solution.
Let \(f(x,y) = xy\) and the constraint \(g(x,y) = 4x^2 + 9y^2 - 32 = 0\text{.}\) Using Lagrange multipliers, \(\nabla f = \lambda \nabla g\text{:}\)
\begin{align*} y \amp = \lambda(8x) \\ x \amp = \lambda(18y) \\ 4x^2 + 9y^2 \amp = 32 \end{align*}
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Multiply the first equation by \(y\) and the second by \(x\text{:}\)
\begin{align*} y^2 \amp = 8\lambda xy \\ x^2 \amp = 18\lambda xy \end{align*}
If \(\lambda = 0\text{,}\) then \(x = 0\) and \(y = 0\text{,}\) which does not satisfy the constraint \(4(0)^2 + 9(0)^2 = 32\text{.}\) Thus, \(\lambda \neq 0\) and \(xy \neq 0\text{.}\) We can divide the modified equations to get:
\begin{align*} 8\lambda xy \cdot x^2 \amp = 18\lambda xy \cdot y^2 \implies 8x^2 = 18y^2 \implies 4x^2 = 9y^2 \end{align*}
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Substitute \(4x^2 = 9y^2\) into the constraint equation:
\begin{align*} 9y^2 + 9y^2 \amp = 32 \\ 18y^2 \amp = 32 \implies y^2 = \frac{16}{9} \implies y = \pm \frac{4}{3} \end{align*}
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Now solve for \(x\text{:}\)
\begin{gather*} 4x^2 = 9\left(\frac{16}{9}\right) = 16 \implies x^2 = 4 \implies x = \pm 2 \end{gather*}
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We have four critical points to check: \(\left(2, \frac{4}{3}\right)\text{,}\) \(\left(2, -\frac{4}{3}\right)\text{,}\) \(\left(-2, \frac{4}{3}\right)\text{,}\) and \(\left(-2, -\frac{4}{3}\right)\text{.}\) Evaluating \(f(x,y) = xy\text{:}\)
\begin{align*} f\left(2, \frac{4}{3}\right) \amp = \frac{8}{3} \amp f\left(-2, -\frac{4}{3}\right) \amp = \frac{8}{3} \\ f\left(2, -\frac{4}{3}\right) \amp = -\frac{8}{3} \amp f\left(-2, \frac{4}{3}\right) \amp = -\frac{8}{3} \end{align*}
The maximum value is \(\frac{8}{3}\) and the minimum value is \(-\frac{8}{3}\text{.}\)
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14.8.13.
\(f(x,y,z) = xy + 2z\, , \qquad x^2 + y^2 + z^2 = 36\)
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Solution.
Let \(g(x,y,z) = x^2 + y^2 + z^2 - 36 = 0\text{.}\) The Lagrange equations \(\nabla f = \lambda \nabla g\) yield:
\begin{align*} y \amp = 2\lambda x \\ x \amp = 2\lambda y \\ 2 \amp = 2\lambda z \implies 1 = \lambda z \\ x^2 + y^2 + z^2 \amp = 36 \end{align*}
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Substitute the second equation into the first:
\begin{gather*} y = 2\lambda(2\lambda y) \implies y = 4\lambda^2 y \implies y(1 - 4\lambda^2) = 0 \end{gather*}
This gives us two cases: either \(y = 0\) or \(4\lambda^2 = 1\text{.}\)
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Case 1: \(y = 0\text{.}\) If \(y = 0\text{,}\) then \(x = 2\lambda(0) = 0\text{.}\) Substituting \(x=0\) and \(y=0\) into the constraint gives:
\begin{align*} 0^2 + 0^2 + z^2 \amp = 36 \implies z = \pm 6 \end{align*}
This gives the points \((0,0,6)\) and \((0,0,-6)\text{.}\)
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Case 2: \(4\lambda^2 = 1 \implies \lambda = \pm \frac{1}{2}\text{.}\) If \(\lambda = \frac{1}{2}\text{,}\) then \(y = x\) and from the third equation, \(z = \frac{1}{1/2} = 2\text{.}\) Substituting these into the constraint:
\begin{align*} x^2 + x^2 + 2^2 \amp = 36 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = \pm 4 \end{align*}
This gives the points \((4,4,2)\) and \((-4,-4,2)\text{.}\)
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If \(\lambda = -\frac{1}{2}\text{,}\) then \(y = -x\) and \(z = \frac{1}{-1/2} = -2\text{.}\) Substituting into the constraint:
\begin{align*} x^2 + (-x)^2 + (-2)^2 \amp = 36 \implies 2x^2 = 32 \implies x = \pm 4 \end{align*}
This gives the points \((4,-4,-2)\) and \((-4,4,-2)\text{.}\)
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Now evaluate \(f(x,y,z) = xy + 2z\) at all these points:
\begin{align*} f(0,0,6) \amp = 0 + 12 = 12 \\ f(0,0,-6) \amp = 0 - 12 = -12 \\ f(4,4,2) \amp = 16 + 4 = 20 \\ f(-4,-4,2) \amp = 16 + 4 = 20 \\ f(4,-4,-2) \amp = -16 - 4 = -20 \\ f(-4,4,-2) \amp = -16 - 4 = -20 \end{align*}
The absolute maximum is \(20\) and the absolute minimum is \(-20\text{.}\)
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14.8.19.

The surface area of a right-circular cone of radius \(r\) and height \(h\) is
\begin{equation*} S = \pi r\sqrt{r^2 + h^2} \end{equation*}
and its volume is
\begin{equation*} V = \frac{1}{3}\pi r^2h \end{equation*}
  1. Determine the ratio \(h/r\) for the cone with given surface area \(S\) and maximum volume \(V\text{.}\)
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  2. What is the ratio \(h/r\) for a cone with given volume \(V\) and minimum surface area \(S\text{?}\)
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  3. Does a cone with given volume \(V\) and maximum surface area \(S\) exist?
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Solution.
Before we begin, note that we can use Lagrange multipliers by finding the gradients of \(V\) and \(S\) with respect to \(r\) and \(h\text{:}\)
\begin{align*} \nabla V \amp = \left\langle \frac{2}{3}\pi rh, \frac{1}{3}\pi r^2 \right\rangle \\ \nabla S \amp = \left\langle \pi\sqrt{r^2+h^2} + \frac{\pi r^2}{\sqrt{r^2+h^2}}, \frac{\pi rh}{\sqrt{r^2+h^2}} \right\rangle = \left\langle \pi\frac{2r^2+h^2}{\sqrt{r^2+h^2}}, \frac{\pi rh}{\sqrt{r^2+h^2}} \right\rangle \end{align*}
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  1. To maximize \(V\) given \(S\text{,}\) we set \(\nabla V = \lambda \nabla S\text{:}\)
    \begin{align*} \frac{2}{3}\pi rh \amp = \lambda \pi\frac{2r^2+h^2}{\sqrt{r^2+h^2}} \\ \frac{1}{3}\pi r^2 \amp = \lambda \frac{\pi rh}{\sqrt{r^2+h^2}} \end{align*}
    Assuming \(r, h \neq 0\text{,}\) we can isolate \(\lambda\) in both equations and set them equal, or simply divide the first equation by the second:
    \begin{align*} \frac{\frac{2}{3}\pi rh}{\frac{1}{3}\pi r^2} \amp = \frac{\lambda \pi\frac{2r^2+h^2}{\sqrt{r^2+h^2}}}{\lambda \frac{\pi rh}{\sqrt{r^2+h^2}}} \\ \frac{2h}{r} \amp = \frac{2r^2+h^2}{rh} \end{align*}
    Cross-multiply to get:
    \begin{align*} 2h^2r \amp = r(2r^2+h^2) \\ 2h^2 \amp = 2r^2+h^2 \implies h^2 = 2r^2 \implies \frac{h^2}{r^2} = 2 \implies \frac{h}{r} = \sqrt{2} \end{align*}
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  2. To minimize \(S\) given \(V\text{,}\) the Lagrange condition is \(\nabla S = \mu \nabla V\text{.}\) Because the vectors must still be parallel, the proportion between the components must be exactly the same as in part (a). Therefore, the algebra will lead to the exact same geometric ratio: \(\frac{h}{r} = \sqrt{2}\text{.}\)
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  3. No, such a cone does not exist. If \(V\) is constant, \(h = \frac{3V}{\pi r^2}\text{.}\) Substitute this into the surface area formula:
    \begin{align*} S \amp = \pi r \sqrt{r^2 + \left(\frac{3V}{\pi r^2}\right)^2} = \sqrt{\pi^2 r^4 + \frac{9V^2}{r^2}} \end{align*}
    If we stretch the cone to be infinitely wide and flat (\(r \to \infty\)), the surface area \(S \to \infty\text{.}\) Thus, there is no maximum surface area for a given volume.
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14.8.23.

Find the point \((x_0, y_0)\) on the line \(4x + 9y = 12\) that is closest to the origin.
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Solution.
We want to minimize the distance to the origin, which is \(d = \sqrt{x^2+y^2}\text{.}\) To make the calculus easier, we will instead minimize the square of the distance, \(f(x,y) = x^2+y^2\text{,}\) subject to the constraint \(g(x,y) = 4x + 9y - 12 = 0\text{.}\)
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Using Lagrange multipliers, \(\nabla f = \lambda \nabla g\text{:}\)
\begin{align*} 2x \amp = \lambda(4) \implies x = 2\lambda \\ 2y \amp = \lambda(9) \implies y = \frac{9}{2}\lambda = 4.5\lambda \end{align*}
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Substitute these into the constraint equation:
\begin{align*} 4(2\lambda) + 9(4.5\lambda) \amp = 12 \\ 8\lambda + 40.5\lambda \amp = 12 \\ 48.5\lambda \amp = 12 \implies \lambda = \frac{12}{48.5} = \frac{24}{97} \end{align*}
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Now plug \(\lambda\) back in to find \(x_0\) and \(y_0\text{:}\)
\begin{align*} x_0 \amp = 2\left(\frac{24}{97}\right) = \frac{48}{97} \\ y_0 \amp = \frac{9}{2}\left(\frac{24}{97}\right) = \frac{108}{97} \end{align*}
The closest point is \(\left(\frac{48}{97}, \frac{108}{97}\right)\text{.}\)
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14.8.29.

Show that the minimum distance from the origin to a point on the plane \(ax + by + cz = d\) is
\begin{equation*} \frac{|d|}{\sqrt{a^2 + b^2 + c^2}} \end{equation*}
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Solution.
Similar to the previous problem, we minimize the square of the distance, \(f(x,y,z) = x^2+y^2+z^2\text{,}\) subject to the plane constraint \(g(x,y,z) = ax + by + cz - d = 0\text{.}\)
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Set up the Lagrange equations \(\nabla f = \lambda \nabla g\text{:}\)
\begin{align*} 2x \amp = a\lambda \implies x = \frac{a\lambda}{2} \\ 2y \amp = b\lambda \implies y = \frac{b\lambda}{2} \\ 2z \amp = c\lambda \implies z = \frac{c\lambda}{2} \end{align*}
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Substitute these into the constraint equation:
\begin{align*} a\left(\frac{a\lambda}{2}\right) + b\left(\frac{b\lambda}{2}\right) + c\left(\frac{c\lambda}{2}\right) \amp = d \\ \frac{\lambda}{2}(a^2 + b^2 + c^2) \amp = d \implies \lambda = \frac{2d}{a^2+b^2+c^2} \end{align*}
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Now substitute \(\lambda\) back to find the coordinates:
\begin{align*} x \amp = \frac{a}{2}\left(\frac{2d}{a^2+b^2+c^2}\right) = \frac{ad}{a^2+b^2+c^2} \\ y \amp = \frac{b}{2}\left(\frac{2d}{a^2+b^2+c^2}\right) = \frac{bd}{a^2+b^2+c^2} \\ z \amp = \frac{c}{2}\left(\frac{2d}{a^2+b^2+c^2}\right) = \frac{cd}{a^2+b^2+c^2} \end{align*}
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Finally, calculate the minimum distance \(D = \sqrt{x^2+y^2+z^2}\text{:}\)
\begin{align*} D \amp = \sqrt{\left(\frac{ad}{a^2+b^2+c^2}\right)^2 + \left(\frac{bd}{a^2+b^2+c^2}\right)^2 + \left(\frac{cd}{a^2+b^2+c^2}\right)^2} \\ \amp = \sqrt{\frac{a^2d^2 + b^2d^2 + c^2d^2}{(a^2+b^2+c^2)^2}} = \sqrt{\frac{d^2(a^2+b^2+c^2)}{(a^2+b^2+c^2)^2}} \\ \amp = \sqrt{\frac{d^2}{a^2+b^2+c^2}} = \frac{|d|}{\sqrt{a^2+b^2+c^2}} \end{align*}
This confirms the distance formula.
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14.8.41.

Find the minimum of \(f(x,y,z) = x^2 + y^2 + z^2\) subject to the two constraints \(x + y + z = 1\) and \(x + 2y + 3z = 6\text{.}\)
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Solution.
Here we have two constraints. Let \(g(x,y,z) = x + y + z - 1 = 0\) and \(h(x,y,z) = x + 2y + 3z - 6 = 0\text{.}\) Using \(\nabla f = \lambda \nabla g + \mu \nabla h\text{:}\)
\begin{align*} 2x \amp = \lambda(1) + \mu(1) \\ 2y \amp = \lambda(1) + \mu(2) \\ 2z \amp = \lambda(1) + \mu(3) \end{align*}
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We can solve this by substituting these expressions into the two constraint equations. First, let’s look at the \(g\) constraint: \(x + y + z = 1 \implies 2x + 2y + 2z = 2\text{.}\) Substitute our gradient equations into this:
\begin{align*} (\lambda + \mu) + (\lambda + 2\mu) + (\lambda + 3\mu) \amp = 2 \\ 3\lambda + 6\mu \amp = 2 \qquad \text{(Equation A)} \end{align*}
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Now for the \(h\) constraint: \(x + 2y + 3z = 6 \implies 2x + 4y + 6z = 12\text{.}\) Substitute our gradient equations:
\begin{align*} (\lambda + \mu) + 2(\lambda + 2\mu) + 3(\lambda + 3\mu) \amp = 12 \\ \lambda + \mu + 2\lambda + 4\mu + 3\lambda + 9\mu \amp = 12 \\ 6\lambda + 14\mu \amp = 12 \qquad \text{(Equation B)} \end{align*}
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We now have a system of two variables. Multiply Equation A by 2:
\begin{align*} 6\lambda + 12\mu \amp = 4 \end{align*}
Subtract this from Equation B:
\begin{align*} (6\lambda + 14\mu) - (6\lambda + 12\mu) \amp = 12 - 4 \\ 2\mu \amp = 8 \implies \mu = 4 \end{align*}
Substitute \(\mu = 4\) back into Equation A to find \(\lambda\text{:}\)
\begin{align*} 3\lambda + 6(4) \amp = 2 \implies 3\lambda = -22 \implies \lambda = -\frac{22}{3} \end{align*}
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Now we can find our coordinates:
\begin{gather*} 2x = -\frac{22}{3} + 4 = -\frac{10}{3} \implies x = -\frac{5}{3} \\ 2y = -\frac{22}{3} + 8 = \frac{2}{3} \implies y = \frac{1}{3} \\ 2z = -\frac{22}{3} + 12 = \frac{14}{3} \implies z = \frac{7}{3} \end{gather*}
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Finally, plug this point into \(f(x,y,z)\) to find the minimum value:
\begin{align*} f\left(-\frac{5}{3}, \frac{1}{3}, \frac{7}{3}\right) \amp = \left(-\frac{5}{3}\right)^2 + \left(\frac{1}{3}\right)^2 + \left(\frac{7}{3}\right)^2 \\ \amp = \frac{25}{9} + \frac{1}{9} + \frac{49}{9} = \frac{75}{9} = \frac{25}{3} \end{align*}
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14.8.45.

The cylinder \(x^2 + y^2 = 1\) intersects the plane \(x + z = 1\) in an ellipse. Find the point on such an ellipse that is farthest from the origin.
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Solution.
We want to maximize the distance to the origin, which means maximizing \(f(x,y,z) = x^2+y^2+z^2\text{.}\) The constraints are \(g(x,y,z) = x^2+y^2 - 1 = 0\) and \(h(x,y,z) = x+z - 1 = 0\text{.}\)
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Set up \(\nabla f = \lambda \nabla g + \mu \nabla h\text{:}\)
\begin{align*} 2x \amp = 2\lambda x + \mu \\ 2y \amp = 2\lambda y \\ 2z \amp = \mu \end{align*}
From the second equation, \(2y(1 - \lambda) = 0\text{,}\) so either \(y = 0\) or \(\lambda = 1\text{.}\)
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Case 1: \(y = 0\text{.}\) Substitute this into the first constraint: \(x^2 + 0^2 = 1 \implies x = \pm 1\text{.}\) If \(x = 1\text{,}\) the second constraint gives \(1 + z = 1 \implies z = 0\text{.}\) This is point \((1, 0, 0)\text{.}\) If \(x = -1\text{,}\) the second constraint gives \(-1 + z = 1 \implies z = 2\text{.}\) This is point \((-1, 0, 2)\text{.}\)
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Case 2: \(\lambda = 1\text{.}\) Substitute into the first equation: \(2x = 2(1)x + \mu \implies \mu = 0\text{.}\) If \(\mu = 0\text{,}\) the third equation gives \(2z = 0 \implies z = 0\text{.}\) Using the second constraint, \(x + 0 = 1 \implies x = 1\text{.}\) Using the first constraint, \(1^2 + y^2 = 1 \implies y = 0\text{.}\) We arrive at the point \((1,0,0)\) again.
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Evaluating the distance squared for our two candidates:
\begin{align*} f(1,0,0) \amp = 1^2 + 0^2 + 0^2 = 1 \\ f(-1,0,2) \amp = (-1)^2 + 0^2 + 2^2 = 5 \end{align*}
The maximum distance squared is 5, so the farthest point from the origin is \((-1, 0, 2)\text{.}\)
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