Table of contents 
Maxima and Minima of Functions of Two Variables 
Functions of Three Variables 
Method of Lagrange Multiplier 
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Reminder
For a function of one variable f(x), we find the local maxima/minima by differentiation. Maxima/ minima occur when f’(x) = 0.
A point where f”(a) = 0 and f”'(a) ≠ 0 is called a point of inflection. Geometrically, the equation y = f(x) represents a curve in the twodimensional (x,y) plane, and we call this curve the graph of the function f(x).
Functions of two variables
Our aim is to generalise these ideas to functions of two variables. Such a function would be written as z = f(x,y)
where x and y are the independent variables and z is the dependent variable. The graph of such a function is a surface in three dimensional space. A simple example might be
z is the height of the surface above a point (x,y) in the XY plane.
For functions z = f(x,y) the graph (i.e. the surface) may have maximum points or minimum points (or both). But for surfaces there is a third possibility  a saddle point.
A point (a,b) which is a maximum, minimum or saddle point is called a stationary point. The actual value at a stationary point is called the stationary value. What we need is a mathematical method for finding the stationary points of a function f(x,y) and classifying them into maximum, minimum or saddle point. This method is analogous to, but more complicated than, the method of working out first and second derivatives for functions of one variable.
Let’s remind about partial derivatives. The sort of function we have in mind might be something like.
f(x,y) = x^{2}y^{3} + 3y + x
and the partial derivatives of this would be
Note that
This is true for any well behaved function. In terms of notation, we will frequently use the other, subscript, notation for partial derivatives:
Finding stationary points
To find the stationary points of f(x,y), work out and set both to zero. This gives you two equations for two unknowns x and y. Solve these equations for x and y (often there is more than one solution, as indeed you should expect. After all, even functions of one variable may have both maximum and minimum points).
Classifying stationary points
The procedure for classifying stationary points of a function of two variables is analogous to, but somewhat more involved, than the corresponding ‘second derivative test’ for functions of one variable. Below is, essentially, the second derivative test for functions of two variables:
Let (a,b) be a stationary point, so that f_{x} = 0 and f_{y} = 0 at (a, b). Then:
Distinguish between these as follows:
if then anything is possible. More advanced methods are required to classify the stationary point properly.
Let's give some idea where the above conditions come from. It is all based on Taylor’s theorem for a function of two variables. Taylor’s theorem for a function of one variable i
For a function of two variables Taylor’s theorem isThe higher order terms can be neglected in straightforward cases. Lets suppose (a,b) is a maximum point. Then f_{x} = 0 and f_{y} = 0 at (a,b) and, because (a,b) is a local maximum the function must x y be smaller at neighbouring points, i.e. when h and k are sufficiently small, f(a + h, b + k) < f(a, b)
But from Taylors theorem, neglecting higher order terms and noting that the first derivative terms are zero at (a.b),
However f(a + h, b + k) < f(a, b), hence
h^{2}f_{xx} + 2hkf_{xy} + k^{2}f_{yy} < 0 at (a,b)
for all small values of h and k. Dividing by k^{2} gives
Let ξ = h/k. Then even though h and k are both small, ξ doesn’t have to be small.
So we have
Thus we have a quadratic expression that is negative for all values of its variable ξ (and so, in particular, has no roots). A few graphs will show that this is only possible if f_{xx}< 0, f_{yy} < 0 and the latter condition is the one to do with having no roots. All these inequalities hold at (a,b).
Similar analysis yields the conditions under which a stationary point is a minimum or saddle point.
Definition
Example 1: Find the maxima and minima of the function
f(x, y) = x^{3} + y^{3}  3x  12y + 20.
We have
f_{x} = 3x^{2}  3, f_{y} = 3y^{2}  12, f_{xx} = 6x, f_{yy} = 6y, f_{xy} = 0.
To determine the critical points, we solve
f_{x} = 0, f_{y} = 0 ⇒ 3x^{2}  3 = 0, 3x^{2}  12 = 0
⇒ x = ± 1, y = ± 2. Hence the critical points are (1,2), (12), (1,2), (1,2).At (1,2),
A = f_{xx} = 6 > 0, B=0, C=12.
AC  B^{2} = f_{xx}. f_{yy}  (f_{xy})^{2} = 6 x 12  0 = 72 > 0.
Thus f has minimum at (1, 2).At (1,2),
AC  B^{2} = 6 * (12)  0 < 0.
Thus f has neither maximum nor minimum at (1, 2).At (1,2),
AC  B^{2} = 6 * 12  0 < 0.
Thus f has neither maximum nor minimum at (1, 2).
At (1,2), A =  6 < 0,
AC  B^{2} = (6)(12)  0 > 0.
Thus f has maximum at (1, 2).
Example 2: Lets work out the stationary points for the function f(x,y) = x^{2} + y^{2 }and classify them into maxima, minima and saddles.
We need all the first and second derivatives so lets work them out. we have
f_{x} = 2x
f_{y} = 2y
f_{xx} = 2f_{yy}= 2
f_{xy} = 0For stationary points we need f_{x} = f_{y} = 0. This gives 2x = 0 and 2y = 0 so that there is just one stationary point, namely (x.y) = (0,0). We now need to classify it.
Nowso it is either a max or a min. But f_{xx} = 2 > 0 and f_{yy} = 2 > 0. Hence it is a minimum. Our conclusion is that this function has just one stationary point (0, 0) and that it is a minimum.
Example:
The first and second order partial derivatives of this function are:
Stationary points are when f_{x} = 0 and f_{y} = 0 and so there is only one stationary point, at (x,y) = (0, 0). Substituting (x.y) = (0,0) into the expressions for f_{xx} , f_{yy} and f_{xy} gives
f_{xx} = 2, f_{yy} = 2, f_{xy} = 0
Therefore
So that (0,0) is either a min or a max. Since f_{xx} < 0 and f_{yy} < 0 it is a maximum.
Example: f(x,y) = 2  x^{2}  xy  y^{2}
For this function
f_{x}=  2x  y
f_{y }= x 2y
f_{xx} = 2
f_{yy}= 2
f_{xy} = 1
For stationary points, 2x  y = 0 and x  2y = 0 so again the only possibility is (x, y) = (0,0). We have
so that (0, 0) is either a max or a min. Since f_{xx} < 0 and f_{yy} < 0, So it is a maximum.
We extend our study of multivariable functions to functions of three variables.
Definition: Function of Three Variables. Let D be a subset of ℝ^{3}. A function f of three variables is a rule that assigns each triple (x, y, z) in D a value w = f(x, y, z) in ℝ . D is the domain of f; the set of all outputs of f is the range.
Note how this definition closely resembles that of Function of two variables.
Example 1: Let . Evaluate f at the point (3, 0, 2) and find the domain and range of f.
As the domain of f is not specified, we take it to be the set of all triples (x, y, z) for which f(x, y, z) is defined. As we cannot divide by 0, we find the domain D is D = {(x, y, z) I x + 2y  z ≠ 0}.
We recognize that the set of all points in ℝ^{3 }that are not in D form a plane in space that passes through the origin (with normal vector (1, 2, 1)).
We determine the range R is ℝ; that is, all real numbers are possible outputs of f. There is no set way of establishing this. Rather, to get numbers near 0 we can let y = 0 and choose z ≈  x^{2}. To get numbers of arbitrarily large magnitude, we can let z ≈ x + 2y.
Level Surfaces
If is very difficult to produce a meaningful graph of a function of three variables. A function of one variable is a curve drawn in 2 dimensions; a function of two variables is a surface drawn in 3 dimensions; a function of three variables is a hypersurface drawn in 4 dimensions.
There are a few techniques one can employ to try to “picture” a graph of three variables. One is an analogue to level curves: level surfaces. Given w = f(x, y, z), the level surface at w = c is the surface in space formed by all points (x, y, z) where f(x, y, z) = c.
Example: Finding level surfaces. If a point source S is radiating energy, the intensity I at a given point P in space is inversely proportional to the square of the distance between S and P. That is, when S = (0, 0, 0).
for some constant k.
Let k = 1; find the level surfaces of I.
If energy (say, in the form of light) is emanating from the origin, its intensity will be the same at all points equidistant from the origin. That is, at any point on the surface of a sphere centered at the origin, the intensity should be the same. Therefore, the level surfaces are spheres.
We now find this mathematically. The level surface at I = c is defined byA small amount of algebra reveals
Given an intensity c, the level surface I = c is a sphere of radius 1/√c, centered at the origin.
Definition: Open Balls, Limit, Continuous
These definitions can also be extended naturally to apply to functions of four or more variables. Theorem also applies to function of three or more variables, allowing us to say that the function is continuous everywhere.
Partial Derivatives and Functions of Three Variables
The concepts underlying partial derivatives can be easily extend to more than two variables. We give some definitions and examples in the case of three variables and one can extend these definitions to more variables if needed.
Definition: Partial Derivatives with Three Variables. Let w = f(x, y, z) be a continuous function on an open set S in ℝ^{3}.
The partial derivative of f with respect to x is :
Similar definitions hold for fy(x, y, z) and fz(x, y, z).By taking partial derivatives of partial derivatives, we can find second partial derivatives of f with respect to z then y, for instance, just as before.
Example. Partial derivatives of functions of three variables.
For each of the following, find f_{x}, f_{y}, f_{z}, f_{x}_{z}, f_{y}_{z}, and f_{z}_{z}.
1. f_{x} = 2xy^{3}z^{4} + 2xy^{2} + 3x^{2}z^{3};
f_{y} = 3x^{2}y^{2}z^{4} + 2x^{2}y + 4y^{3}z^{4};
f_{z} = 4x^{2}y^{3}z^{3} + 3x^{3}z^{2} + 4y^{4}z^{3};
f_{xz} = 8xy^{3}z^{3} + 9x^{2}z^{2};
f_{yz} = 12x^{2}y^{2}z^{3} + 16y3z^{3};
f_{zz} = 12x^{2}y^{3}z^{2} + 6x^{3}z + 12y^{4}z^{2}
2. f_{x} = sin(yz);
f_{y} = xz cos(yz);
f_{z} = xy cos(yz);
f_{x}_{z }= y cos(yz);
f_{y}_{z }= x cos(yz)  xyz sin(yz);
f_{zz }=  xy^{2}sin (xy)
Higher Order Partial Derivatives
We can continue taking partial derivatives of partial derivatives of partial derivatives of....; we do not have to stop with second partial derivatives.
We do not formally define each higher order derivative, but rather give just a few examples of the notation.
Example: Higher order partial derivatives.
1. Let f(x, y) = x^{2}y^{2} + sin(xy). Find f_{xxy}, and f_{yxx}.
2. Let f(x, y, z) = x^{3}e^{xy} + cos(z). Find f_{xyz}.
 To find f_{xxy} we first find f_{x}, then f_{xx}, then f_{xxy}:
f_{x} = 2xy^{2} + y cos(xy)
f_{xx} = 2y^{2}  y^{2} sin<xy)
f = 4y  2y sin(xy)  xy^{2} cos(xy).
To find f_{yxx} we first find f_{y} then f_{yx} then f_{yxx} :
f_{y} = 2x^{2}y + x cos(xy)
f_{yx}= 4xy + cos(xy)  xy sin(xy)
f_{yxx}= 4y  y sin(xy)  (y sin(xy) + xy^{2} cos(xy))
= 4y  2y sin(xy)  xy^{2} cos(xy).
Note how = f_{xxy} = f_{yxx}. To find f , we find f_{x}, then f_{xy}, then f_{xyz} :
f_{x} = 3x^{2}e^{xy} + x^{3}ye^{xy}
f_{xy} = 3x^{3}e^{xy} + x^{3} e^{xy} + x^{4}ye^{xy} = 4x^{3}e^{xy} + x^{4}ye^{xy}
f_{xyz }=0
In the previous example we saw that f_{xxy}= f_{yxx} ; this is not a coincidence. While we do not state this as a formal theorem, as long as each partial derivative is continuous, it does not matter the order in which the partial derivatives are taken. For instance, f_{xxy} = f_{xy}_{z} = f_{yxx.}
Differentiability of Functions of Three Variables
The definition of differentiability for functions of three variables is very similar to that of functions of two variables. We again start with the total differential.
Definition : Total Differential. Let w = f(x, y, z) be continuous on an open set S. Let dx, dy and dz represent changes in x, y and z, respectively. Where the partial derivatives f_{x}, f_{y} and f_{z} exist, the total differential of w is
dz = f_{x}(x, y, z)dx + f_{y}(x, y, z)dy + f_{z}(x, y, z)dz .
This differential can be a good approximation of the change in w when w = f(x, y, z) is differentiable.
Definition : Multivariable Differentiability. Let w = f(x, y, z) be defined on an open ball B containing (x_{0}, y_{0}, z_{0}) where f_{x}(x_{0}, y_{0}, z_{0}), f_{y}(x_{0}, y_{0}, z_{0}) and f_{z}(x_{0}, y_{0}, z_{0}) exist. Let dw be the total differential of w at (x_{0}, y_{0}, z_{0}). Let Δw = f(x_{0} + dx, y_{0} + dy, z_{0} + dz)  f(x_{0}, y_{0}, z_{0}), and let E_{x}, E_{y} and E_{z} be functions of dx, dy and dz such that
Δw = dw = E_{x}dx + E_{y}dy + E_{z}dz
Just as before, this definition gives a rigorous statement about what it means to be differentiable that is not very intuitive. We follow it with a theorem similar to Theorem.
Theorem: Continuity and Differentiability of Functions of Three Variables.
Let w = f(x, y, z) be defined on an open ball B containing (x_{0}, y_{0}, z_{0}).
This set of definition and theorem extends to functions of any number of variables. The theorem again gives us a simple way of verifying that most functions that we encounter are differentiable on their natural domains.
This section has given us a formal definition of what it means for a functions to be "differentiable," along with a theorem that gives a more accessible understanding. The following sections return to notions prompted by our study of partial derivatives that make use of the fact that most functions we encounter are differentiable.
Maxima & Minima of Three Variables Functions
Let u = f(x, y, z) be a function of three variables x, y and z, then we can find the maximum and minimum value of the function as like the two variables.
Suppose f(x, y, z) is a given function of three independent variables x, y and z.
Example: Discuss the maximum and minimum value of
u = x^{2} + y^{2} + z^{2} + x  2z  xy.
As given
u = x^{2} + y^{2} + z^{2} + x  2z  xy, then
u_{x} = 2x  y + 1 = 0, u_{y} = x + 2y = 0, u_{z} = 2z  2 = 0.
On solving these equations we get x = 2/3, y = 1/3, z = 1. Therefore is only critical point where function will be either maximum or minimum. Now we will decide it.Now the values of following derivatives A = 2, B = 2, C = 2, F = 0, G = 0, H = 1
Now we have
Since all these A, A_{1}, and A_{2} are positive, we have a minimum of u at the point
Example: Show that the point such that the sum of the sum of the squares of its distances from n given points shall be minimum, is the center of the mean position of the given points. Let the n given points be (a_{1}, b_{1}, c_{1}), (a_{2}, b_{2}, c_{3}), ... (a_{n}, b_{n}, c_{n}) and let (x, y, z) be the coordinate of the required point.
If u denotes the sum of the squares of the distances of (x, y, z) from the n given points, then
For maximum or a minimum of u, we must have
.....(2)
...... (3)
...... (4)
Solving these above equations, we get
Now A = 2n, B = 2n, C = 2n, F = G = H = 0
We have A = 2n_{1}, A_{1} = 4n^{2}, A_{2} = 8n^{3}_{.} Since there three expression are all positive, u is minimum when
So u is minimum when the point (x, y, z) is the centre of the mean position of the n given points.
Example 1: Find the range of
x + 2y  4z ≠ 0; the set of points of ℝ^{3} NOT in the domain form a plane through the origin. Range: ℝ
Example 2: Find the range of
x^{2} + y^{2} + z^{2} ≠ 1; the set of points in ℝ^{3} NOT in the domain form a sphere of radius 1.
Range: (∞, 0) U [1, ∞)
Example 3: Find the range of
z ≥ x^{2}  y^{2}; the set of points in ℝ^{3} above (and including) the hyperbolic paraboloid z = x^{2}  y^{2}. Range: [0, ∞)
The method of Lagrange multipliers allows us to maximize or minimize functions with the con straint that we only consider points on a certain surface. To find critical points of a function f(x, y, z) on a level surface g(x, y, z) = C (or subject to the constraint g(x, y, z) = C), we must solve the following system of simultaneous equations:
∇f(x,y,z) = λ∇g(x,y,z)
g(x, y, z) = C
Remembering that ∇f and ∇g are vectors, we can write this as a collection of four equations in the four unknowns x, y, z, and λ:
f_{x}(x,y, z) = λg_{x}(x, y, z)
f_{y}(x,y, z) = λg_{y}(x, y, z)
f_{z}(x,y, z) = λg_{z}(x, y, z)
g(x, y, z) = C
The variable λ is a dummy variable called a “Lagrange multiplier”; we only really care about the values of x, y, and z.
Once you have found all the critical points, you plug them into f to see where the maxima and minima of the function occurs. The critical points where f is greatest are called points of maxima and the critical points where f is smallest are called points of minima.
Example 1: Use the method of Lagrange multipliers to maximize f(x, y) = 3xy.
subject to the constraint x + y = 8.
First let f(x, y) = 3xy and g(x, y) = x + y  8.
then defined a new function F as
F(x, y, z) = f(x, y)  λg(x, y)
= 3xy  λ(x + y  8)
To find the critical numbers of F, begin by finding the partial derivatives of F with respect to x, y and λ. Then, set of partial derivatives equal to zero.F_{x}(x, y, λ) = 3y  λ ⇒ 3y  λ = 0 ...(1)
F_{y}(x, y, λ) = 3x  λ ⇒ 3x  λ = 0 ...(2)
F_{z}(x, y, λ) = x  y + 8 ⇒ xy + 8 = 0 ...(3)
Solving for λ in the first equation (1) gives
3y  λ = 0 ⇒ λ = 3y
Substituting for λ in the second (2) equation
3x  3y = 0 ⇒ 3x  3y or x = y
Now, substitute the value of x in equation (3).
 y  y + 8 = 0
2y = 8
y = 4
Using this value, we can conclude that the critical values are x = 4, y = 4. Which implies that the maximum is
f(x, y) = 3xy
= 3(4)(4) = 48.
Example 2: Find the maximum of V = xyz subject to the constraint 6x + 4y + 3z  24 = 0
First, let f(x, y, z) = xyz and g(x, y, z) = 6x + 4y + 3z  24.
Then, define a new function F as F(x, y, z, λ) = f(x, y, z)  λg(x, y, z)
= xyz  X(6x + 4y + 3z  24).To find the critical numbers of F, begin by finding the partial derivatives of F with respect to x, y, z, and λ. Then, set the partial derivatives equal to zero.
F_{x}(x, y, z, λ ) = yz  6λ ⇒ yz  6λ = 0
Fy(x, y, z, λ) = xz  4λ ⇒ xz  4λ = 0
Fz(x, y, z, λ) = xy  3λ ⇒ xy  3λ = 0
F. (x, y, z, λ) =  6x  4y  3z + 24 ⇒ 6x  4y  3z + 24 = 0
Solving for λ in the first equation producesSubstituting for λ in the second and third equations produces the following.
Next, substituting for y and z in the equation F_{x}(x, y, z, λ) = 0 and solve for x.
Using this xvalue, you can conclude that the critical values are x = 4/3, y = 2 and z = 8/3. which implies that the maximum is V = xyz. Write objective function
Example 3: Minimize: f(x, y) = x^{2} + y^{2}
Subject to : 2x + 6y = 2000
f(x, y) = x^{2} + y^{2}
g(x, y) = 2x + 6y  2000 = 0.L(x, y, λ) = x^{2} + y^{2}  λ (2x + 6y  2000)
L_{x}(x, y, λ) = 2x  2λ ⇒ 2x  2λ = 0 ...(1)
L_{y}(x , y, λ) = 2y  6λ ⇒ 2y  6λ = 0 ...(2)
L_{λ}(x, y, λ) =  2 x  6y + 2000 ⇒  2 x  6y + 2000 = 0 ...(3)Solving for X
2x — 2λ = 0 => λ = x
Putting value of λ in (2).
2y + 6x = 0 => 6x = 2y, y = 3xNow Substituting y in (3)
2x  18x + 2000 = 0
⇒ 20x + 2000 = 0
x = 100
Using the value of x we can conclude that the critical points are x = 100, y = 300.
Therefore the min value in f(100, 300) = (100)^{2} + (300)^{2}= 10000 + 90000 = 1,00,000
Example 4: Find the maximum and minimum values of f(x, y) = 81x^{2} + y^{2} subject to the constraint 4x^{2} + y^{2} = 9.
Let f(x, y) = 81 x^{2} + y^{2}
g(x, y) = 4x^{2} + y^{2}  9
Then define a new function F as
F = f(x, y)  λg(x, y)
= 81 x^{2} + y^{2}  λ(4x^{2} + y^{2}  9)
F_{x}(x, y, λ) = 162x  8λx
⇒ 162x  8λx ..... (1)
F_{y}(x, y, λ) = 2y  2yλ
⇒ 2y  2yλ = 0 ...(2)
F_{λ} (x, y, λ) =  4x^{2}  y^{2} + 9
⇒ 4x^{2} y^{2} + 9 = 0 ...(3)
Solve for λ Let’s start with the second one.
2y = 2yλ
⇒ λ = 1 or y = 0
Let y = 0, then in this caseTherefore we get two points that are absolute extrema
Now Let λ = 1, in this case equation (1) reduces to162x = 8x ⇒ 154x = 0 ⇒ x = 0
under this assumption we must have x = 0. Now substitute in (3) we get
y^{2} = 9 ⇒ y = ± 3
So this gives us two more points that are absolute extreme
(0, 3) and (0, 3)
Now, we have four points that are absolute extrema to determine the absolute extrema we need to evaluate the function at each of these pointsThe absolute maximum is 729/4 = 182.25 which occurs at
The absolute minimum is which occurs at (0, 3) and (0, 3).
Example 5 : Find the maximum values of f(x, y) = 8x^{2} 2y subject to the constraint x^{2}+ y^{2}=1.
Given f(x, y) = 8x^{2}  2y
g(x, y) = x^{2} + y^{2}  1
F(x, y, λ) = f(x, y)  λg(x, y) = 8x^{2}  2y  λ(x^{2} + y^{2} I)
F_{x} = 16x  2λx ⇒ 16x  2λx = 0 ...(1)
F_{y} = 2  2λy ⇒ 2  2λy = 0 ...(2)
F =  x^{2}  y^{2} + 1 ⇒  x^{2}  y^{2} + 1 = 0
Solve for λ
by (1) 16x  2λx = 0
⇒ x = 0 or λ = 8
Let x = 0 then by (3), we get
y^{2} = 1 ⇒ y = ±1
Therefore, we get two points that are absolute extrema (0, 1) and (0, 1) Now assume λ = 8. In this case, we can plug the value in equation (2) we get.
2 = 16y ⇒ y = 1/8
we now plug the value of y in equation (3)
this gives two more points that are absolute extrema.Now
f(0,  1 ) = 2, f(0,1) =  2The absolute maximum is 65/8 = 8.125 occurs at
Example 6: Find the minimum values of f(x, y, z) = xyz subject to the constraint x + 9y^{2} + z^{2} = 4. Assume that x > 0.
f(x, y, z) = xyz
g(x, y, z) = x + 9y^{2} + z^{2} = 4, x ≥ 0
F(x, y, z, λ) = f(x, y, z)  λg(x, y, z)
= xyz  λ(x + 9y^{2} + z^{2}  4)
F_{x} = yz  λ ⇒ yz  λ = 0 ...(1)
F_{y} = xz  18λy ⇒ xz  18λy = 0 ...(2)
F_{z} = xy  2λz ⇒ xy  2λz = 0 ...(3)
F_{λ} = x  9y^{2}  z^{2} + 4 ⇒ x^{2} + 9y^{2} + z^{2}  4 = 0 ...(4)
Now multiplying equation (1) by x, equation (2) by y and equation (3) by z.
We getxyz = λx ...(5)
xyz = 18λy^{2} ...(6)
xyz = 2λz^{2} ...(7)
by equation (5) and (6)
xλ = 18λy^{2} (x  18y^{2})λ = 0
⇒ x = 18y^{2} or λ = 0
by equation (6) and (7)18y^{2}λ = 2z^{2}λ ⇒ (18y^{2}  2z^{2})λ = 0
⇒ z^{2} = 9y^{2} or λ = 0
Now we have two possibilities Either λ = 0 or we have x = 18y^{2} and z^{2} = 9y^{2}.
Let λ = 0, then by equation (1), (2) and (3) we haveyz = 0 ⇒ y = 0 or z = 0 ...(8)
xz = 0 ⇒ x = 0 or z = 0 ...(9)
xy = 0 ⇒ x = 0 or y = 0 ...(10)
from this we noticed that we can’t have all three of the variables be zero but we could have two of then be zero. So this leads to the following two cases that we can plug into the constraint t find the value of the third variable.
y = 0, x = 0 ⇒ z^{2} = 4 => z = ±2
y = 0,z = 0:x = 4
So this gives us the following three absolute extrema
(0, 0, 2), (0, 0, 2) and (4, 0, 0)
Now, let’s check the second possibility from equation (8) z = 0. In this case the (9) equation while be 0 = 0 and so will not be any use. Then (10) however, has the possibilities of x = 0 or y = 0.
z = 0, x = 0 : 9y^{2} = 4 ⇒ y = ± 2/3
This leads to two more absolute extrema
we had another possibility x = 18y^{2} and z^{2} = 9y^{2}
In this case we can plug each of these directly into the constraint to get the following
18y^{2} + 9 y^{2} + 9y2 = 36y^{2} = 4 ⇒ y = ± 1/3
NowTherefore we get the following four absolute extrema
Now
The absolute minimum is 2/3 which occurs at
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