**Method of Lagrange Multipliers:**To find the maximum and minimum values of \(f(x,y,z)\) subject to the constraint \(g(x,y,z) = k\), assuming that these extreme values exist and \(\nabla g \not = \vec{0} \) on the surface \(g(x,y,z) = k\):

- Find all values of \(x,y,z,\) and \(\lambda\) such that

$$

\nabla f(x,y,z) = \lambda \nabla g(x,y,z)

$$

and

$$

g(x,y,z) = k

$$ - Evaluate \(f\) at all the points \((x,y,z)\) that result from step \((1)\). The largest of these values is the maximum value of \(f\); the smallest is the minimum value of \(f\).

**Strategy for Optimizing \(f(x,y)\) Subject to the Constraint \(g(x,y) \leq c \)**

- Find all points in the region \(g(x,y) < c\) where \(\nabla f = 0\) or is undefined.
- Use Lagrange multipiers to find the local extrema of \(f\) on the boundary \(g(x,y) = c\).
- Evaluate \(f\) at the points found in the previous two steps and compare the values.

Find the maximum and minimum values of

$$

f(x,y) = x^2 + y^2

$$

subject to the constraint \(xy = 1\)

Find all values of \(x,y,z,\) and \(\lambda\) such that

$$

\nabla f (x,y) = \lambda \nabla g(x,y)

$$

where \(g(x,y) = xy = 1\).

That is, solve the system

$$

\left \lbrace

\begin{aligned}

2x & = \lambda y \\

2y & = \lambda x \\

xy & = 1

\end{aligned} \right.

$$

Multiplying the first equation by \(x\) and the second by \(y\) reduces the system to

$$

\left \lbrace

\begin{aligned}

x^2 & = y^2 \\

xy & = 1

\end{aligned} \right.

$$

If \(y = 0\) then \(xy \not = 1\), a contradiction.

So \(y \not = 0\) and therefore we can solve for \(x\) in the second equation. Substituting this back into the first equation, reduces the system to

$$

\frac{1}{y^2} = y^2 \Rightarrow y = \pm 1

$$

Thus the points which satisfy the lagrange conditions are:

- \((1,1)\)
- \((-1,-1)\)

Check that this is the case by substituting these points back into the original system. That is, check that \((1,1)\) and \((-1,-1)\) satisfy

$$

\left \lbrace

\begin{aligned}

2x & = \lambda y \\

2y & = \lambda x \\

xy & = 1

\end{aligned} \right.

$$

for some \(\lambda \not = 0\).

$$

f(x,y,z) = 2x+6y+10z

$$

subject to the constraint \(x^2 + y^2 + z^2 = 35\).

Find all values of \(x,y,z,\) and \(\lambda\) such that

$$

\left \lbrace

\begin{aligned}

\nabla f (x,y) & = \lambda \nabla g(x,y) \\

x^2 + y^2 + z^2 & = 35

\end{aligned} \right.

$$

where \(g(x,y) = x^2 + y^2 + z^2 = 35\).

That is, solve the system

$$

\left \lbrace

\begin{aligned}

2 & = 2\lambda x \\

6 & = 2\lambda y \\

10 & = 2 \lambda z \\

x^2 + y^2 + z^2 & = 35

\end{aligned} \right.

$$

Multiplying the first equation by \(yz\), the second by \(xz\), and the third equation by \(xy\) reduces the system to

$$

\left \lbrace

\begin{aligned}

2yz & = 2\lambda xyz \\

6xz & = 2\lambda xyz \\

10xy & = 2 \lambda xyz \\

x^2 + y^2 + z^2 & = 35

\end{aligned} \right.

$$

which is equivalent to

$$

\left \lbrace

\begin{aligned}

2yz & = 6xz\\

6xz & = 10xy\\

x^2 + y^2 + z^2 & = 35

\end{aligned} \right.

$$

which is equivalent to

$$

\left \lbrace

\begin{aligned}

2z(y- 3x)& =0 \\

6xz & = 10xy\\

x^2 + y^2 + z^2 & = 35

\end{aligned} \right.

$$

That is, either \(z = 0\) or \(y = 3x\).

- If \(z = 0\), then

$$

\left \lbrace

\begin{aligned}

0 & = 10xy\\

x^2 + y^2 & = 35

\end{aligned} \right.

$$

So, either \(x = 0\) or \(y = 0\) but not both since then the second equation won’t be satisfied.- If \(x = 0\), then \(y = \pm \sqrt{35}\).
- If \(y = 0\), then \(x = \pm \sqrt{35}\).

Therefore \(f\) has possible extreme values at \((\pm \sqrt{35},0,0)\) and \((0,\pm \sqrt{35},0)\).

- If \(y = 3x\), then

$$

\left \lbrace

\begin{aligned}

6x(z -5x) & = 0\\

x^2 + y^2 + z^2 & = 35

\end{aligned} \right.

$$

So, either \(x = 0\) or \(z =5x\).- If \(x = 0\), then \(y = 0\) and \(z =\pm \sqrt{35}\).
- If \(z = 5x\), then

$$

x^2 + (3x)^2 + (5x)^2 = 35

$$

So, \(35x^2 = 35\). So, \(x = \pm 1\).

Therefore \(f\) has possible extreme values at \((1,3,5)\) and \((-1,-3,-5)\)

Evaluate \(f\) at these six points, the largest is the maximum value of \(f\) and the smallest is the minimum value of \(f\).