Lagrange Multipliers

All warfare is based on deception, Sun tzu, The Art of War.

Recall

• A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x² + y², e^x+y, etc.

• Partial derivatives are derivatives of a function of multiple variables, say f(x₁, x₂, ···, x_n) with respect to one of those variables, with the other variables held constant. They measure how the function changes as one variable changes, while keeping the others constant. $\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$

Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x₀, y₀) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y₀. Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x₀

• $\frac{dw}{ds}\bigg|_{\vec{u}}$ is the directional derivative in the direction of ${\vec{u}}$, the rate of change of the function w as we move in the direction of $\vec{u}$.

$\frac{dw}{ds}\bigg|_{\vec{u}} = ∇w·\vec{u} = |∇w|·|\vec{u}|cos(θ) = |∇w|·cos(θ)$ where θ is the angle between the gradient and the given unit vector.

• The directional derivative is maximized when cos(θ) = 1 ↭ θ = 0, that means, $\vec{u}$ is in the same direction as the gradient ∇w, $\vec{u}=dir(∇w).$

The gradient is the direction in which the function increases fastest (the direction of the steepest ascent) at a given point, and |∇w| = $\frac{dw}{ds}\bigg|_{\vec{u}=dir(∇w)}$.

The directional derivative is minimized when cos(θ) = -1 ↭ θ = 180 ↭ $\vec{u}$ is in the opposite direction of the gradient ∇w. Futhermore, $\frac{dw}{ds}\bigg|_{\vec{u}} = 0$ ↭ cos(θ) = 0 ↭ θ = 90° ↭ $\vec{u}$ ⊥ ∇w.

• The Chain Rule. Let w = f(x, y, z) where x = x(t), y = y(t), and z = z(t) are functions of another variable t, then the derivate of f with respect to t is given by $\frac{dw}{dt} = f_x\frac{dx}{dt} +f_y\frac{dy}{dt}+f_z\frac{dz}{dt}, \frac{dw}{dt} = w_x\frac{dx}{dt} +w_y\frac{dy}{dt}+w_z\frac{dz}{dt} = ∇w·\frac{d\vec{r}}{dt}~$ where ∇w = ⟨w_x, w_y, w_z⟩ and $\frac{d\vec{r}}{dt}=⟨\frac{d\vec{x}}{dt}, \frac{d\vec{y}}{dt}, \frac{d\vec{z}}{dt}⟩$

• The gradient vector is defined as follows: ∇w = ⟨w_x, w_y, w_z⟩ = $⟨\frac{\partial w}{\partial x}, \frac{\partial w}{\partial y}, \frac{\partial w}{\partial z}⟩$. And $\frac{d\vec{r}}{dt}$ is the vector of derivatives of x, y, and z with respect to t: $\frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}⟩$.

The gradient vector is a vector that points in the direction of the steepest increase of the function at a given point. It is perpendicular (orthogonal) to the level surfaces of the function. This means that the gradient points in the direction of the steepest ascent or increase, and there is no change in the function’s value along the level surface, making the gradient perpendicular to it.

Lagrange Multipliers

Lagrange multipliers are a powerful method for finding the local maxima and minima of a function, say f(x, y, z), where the variables involved x, y, and z are not entirely independent, but instead they are constraint by an equation, g(x, y, z) = c. This method leverages the insight that at points where the function f(x, y, z) achieves an extremum (either a maximum or a minimum) subject to additional requirement or constraint g(x, y, z) = c, the gradient of f must be parallel to the gradient of g.

This relationship can be expressed mathematically as: ∇f = λ∇g where λ is the Lagrange multiplier.

Here, λ is known as the Lagrange multiplier. The gradients ∇f and ∇g represent the directions of the steepest ascent for the functions f and g, respectively. The condition that these gradients are parallel implies that at an extremum, the function f cannot increase or decrease in any direction that would remain within the constraint g(x, y, z) = c.

Solved exercises

• Finding the Closest Point to the Origin on a Hyperbola

Let’s apply the method of Lagrange multipliers to a specific problem: finding the point on the hyperbola defined by the equation xy = 3 that is closest to the origin.

Our goal is to minimize the distance from a point (x, y) on the hyperbola to the origin (0, 0). The distance from (x, y) to the origin is given by the function: f(x, y) = $\sqrt{x^2+y^2}$.

However, because the square root function is monotonically increasing, minimizing f(x, y) is equivalent to minimizing its square: f(x, y) = x² + y² (Figure C and D).

The constraint is that the point must lie on the hyperbola g(x, y) = xy = 3. 💡It is like finding the highest or lowest point on a mountain or hill subject to the fact you can only walk along a trail.

Observing at the figure, at the minimum, the level curve of f is tangent to the parabola g. Lagrange multipliers involve finding points (x, y) where the level curves of f and g are tangent, and this is due to the fact that gradients are perpendicular to the contour lines of their respective functions. Therefore, the gradient of f is parallel to the gradient of g, expressed as ∇f // ∇g, meaning they are proportional to each other ↭ ∇f = λ∇g where λ is the Lagrange multiplier.

The gradient of a function at $\vec{a}$ represents the direction in which f increases at the greatest rate so, if one writes down a contour diagram with each contour representing a level set of f (levels of equal value), ∇f($\vec{a}$) represents the direction from $\vec{a}$ where the contours are least sparse.

In Calculus 1, to find the local maxima and minima of a function f(x), we look for points where the derivative f′(x) = 0. At these points, the tangent line to the curve y = f(x) is horizontal, indicating a potential extremum (critical point), e.g., f(x) = x² +4x +4, f’(x) = 2x + 4 = 0 ⇒x = -2 and we need to verify it by the second derivative test.

When finding a minimum or maximum of a function f(x, y, z) subject to a constraint g(x, y, z) = c, we look for points where f does not increase or decrease when moving along the level curve or surface defined by g(x, y, z) = c. Essentially, the rate of change of f in any direction $\vec{a}$ tangent to the level curve g = c must be zero at a constrained extremum. This is because if it were not zero, one could move slightly in the direction of increase or decrease while staying on the level curve or surface g = c, indicating that the current point is not actually an extremum. Mathematically, this is represented as $\frac{df}{ds}\bigg|_{\vec{a}} = 0$ (Figure E)

$\frac{df}{ds}\bigg|_{\vec{a}} = ∇f·\vec{a} = 0$ ⇒ $\vec{a}$ is perpendicular to $∇f$, $\vec{a} ⊥ ∇f$ ⇒ ∇f is perpendicular to the level set of g = c. Since ∇g is also perpendicular to the level set of g = c, it follows ∇f || ∇g.

When extending Calculus 1’s critical points to functions of multiple variables, we seek points where the gradient ∇f is zero to find local extrema. However, when there are constraints, such as in the case of Lagrange multipliers, we look for points where the gradient of f is parallel to the gradient of the constraint function g. This parallelism reflects that the rate of change of f in the direction allowed by the constraint is zero, similar to how the derivative is zero at extrema in one-dimensional calculus.

Therefore, our problem is reformulated as a system of equations,

$\begin{cases} \frac{∂f}{∂x} = λ\frac{∂g}{∂x} \\ \frac{∂f}{∂y} = λ\frac{∂g}{∂y} \\ g = c \end{cases}$

This yields critical points that need to be evaluated to determine if they are maxima, minima, or saddle points, taking into account the constraint g = c. The method does not tell whether a critical point is a minimum, maximum, or saddle point.

$\begin{cases} f_x = λg_x \\ f_y = λg_y \\ g = c\end{cases}$

In our particular example, f(x, y) = x^2 + y^2, g = xy,

$\begin{cases} 2x = λy \\ 2y = λx \\ xy = 3\end{cases}$

$\begin{cases} 2x - λy = 0 \\ λx -2y = 0 \\ xy = 3\end{cases}$

$(\begin{smallmatrix}2 & -λ\\ λ & -2\end{smallmatrix})(\begin{smallmatrix}x\\ y\end{smallmatrix}) = (\begin{smallmatrix}0\\ 0\end{smallmatrix})$.

The trivial solution (0, 0) does not satisfy the constraint (0·0 ≠ 3). We have other solutions only if det(M) = 0 ↭ $|\begin{smallmatrix}2 & -λ\\ λ & -2\end{smallmatrix}|=-4+λ^2=0 ↭ λ^2 = 4 ↭ λ = ± 2$

1. If λ = 2 ⇒ 2x -2y = 0 ⇒º x = y, x² = 3 ⇒ There are two solutions: $(\sqrt{3},\sqrt{3})$ or $(-\sqrt{3},-\sqrt{3})$.
2. If λ = -2, 2x +2y = 0, x = -y, -x² = 3 ⇒ There are no solutions.

The method of Lagrange multipliers has provided us with potential extremum points, but it does not tell us whether these points correspond to a minimum, maximum, or saddle point. In this particular example, since we’re dealing with a distance minimization problem, both points $(\sqrt{3},\sqrt{3})$ and $(-\sqrt{3},-\sqrt{3})$ are minima because they yield the same minimum distance from the origin on the hyperbola.

• Find the dimensions of the largest rectangular box in the first octant with three faces in the coordinates planes and one vertex in the plane 2x + 3y + 5z = 90. What is the maximum volume.

We need to find the dimensions x, y, and z of the largest rectangular box that lies in the first octant (where x, y, and z are all positive) with three of its faces on the coordinate planes xy-plane, yz-plane, and zx-plane, and one vertex on the plane 2x + 3y +5z = 90. The goal is to maximize the volume of this box.

The volume of the box is given by the function f(x ,y, z) = xyz, and it must satisfy the constraint g(x, y, z) =2x + 3y + 5z −90 = 0.

1. Formulate the System Using Lagrange Multipliers. To solve this problem, we use the method of Lagrange multipliers. This method helps us find the extrema of a function subject to a constraint. It involves introducing a new variable, λ (the Lagrange multiplier), and asserts that at an extremum of f(x, y) subject to the constraint g(x, y) =0, the gradients of f and g must be parallel: ∇f = λ∇g. This gives us the following system of equations: $\begin{cases} f_x = λg_x \\ f_y = λg_y \\ g = 0\end{cases}$
2. Compute the Gradients. Let’s compute the partial derivatives f and g: ∇f = ⟨yz, xz xy⟩, ∇g = ⟨2, 3, 5⟩.
3. Set up the equations. Using the gradients calculated above, we set up the system of equations based on the relationship ∇ f= λ∇g: $\begin{cases} yz = 2λ (i) \\ xz = 3λ (ii) \\ xy = 5λ (iii)\\2x + 3y +5 = 90 (iv)\end{cases}$
4. Solve the system of equations. From equation (i), $λ = \frac{yz}{2}$. From equation (ii), $λ = \frac{xz}{3}$. From equation (iii), $λ = \frac{xy}{5}$. As we want to maximize the volume, I do not need to consider z = 0 or y = 0. Equate the first and second expressions: $\frac{yz}{2} = \frac{xz}{3} ⇒ \frac{y}{2} = \frac{x}{3} ⇒ y = \frac{2}{3}x$. Equate the first and third expressions: $\frac{yz}{2} = \frac{xy}{5} ⇒\frac{z}{2} = \frac{y}{5} ⇒ z = \frac{2}{5}x$
5. Substitute back into the constraint: 2x + 3y + 5 = 90 ↭ $2x +3·\frac{2}{3}x + 5\frac{2}{5}x = 90 ↭[\text{Simplify the equation:}] 2x + 2x +2x = 90 ↭ 6x = 90 ↭ x = \frac{90}{6} = 15 ⇒[\text{Now, substitute x = 15 back into the expressions for y and z:}] y = \frac{2}{3}15 = \frac{30}{3} = 10, z = \frac{2}{5}x = \frac{2}{5}15 = \frac{30}{5} = 6$.
6. Calculate the Maximum Volume. Finally, to find the maximum volume, we substitute the values of x, y, and z into the volume function f(x, y, z)= xyz: f(15, 10, 6) = 15·10·60 = 90 cubic units.

• Calculate maximum/minimum of f(x, y) = xy+1, with the constraint: g(x, y) = x²+y²-1 = 0

In this problem, we are tasked with finding the maximum and minimum values of the function f(x, y) = xy +1 given that x and y are constrained by the equation g(x, y) = x²+y²-1 = 0. This constraint represents a circle with a radius of 1 centered at the origin. To solve this problem, we will use the method of Lagrange multipliers, which helps find extrema of functions subject to constraints.

1. Formulate the System Using Lagrange Multipliers. The method of Lagrange multipliers involves introducing a new variable, λ (the Lagrange multiplier), and asserts that at an extremum of f(x, y) subject to the constraint g(x, y) =0, the gradients of f and g must be parallel: ∇f = λ∇g. This gives us the following system of equations: $\begin{cases} f_x = λg_x \\ f_y = λg_y \\ g = 0\end{cases}$
2. Compute the Gradients. Let’s compute the partial derivatives f and g: ∇f = ⟨y, x⟩, ∇g = ⟨2x, 2y⟩.
3. Set up the equations. Using the gradients calculated above, we set up the system of equations based on the relationship ∇ f= λ∇g: $\begin{cases} y = λ2x (i) \\ x = λ2y (ii) \\ x^2 + y^2 = 1\end{cases}$
4. Solve the system of equations. From equation (i), y = λ2x. From equation (ii), x = 2λy. We can substitute x from Equation (ii) into Equation (i): $y = λ2(λ2y)=[(ii)]4λ^2y$. There are two solutions: y = 0⇒[By ii] x = 0, but (0, 0) does not satisfy our constrain x²+y²-1 = 0 (0²+0²-1 = -1 ≠ 0) or y ≠ 0⇒[Dividing by y] $1 = 4λ^2⇒λ^2 = \frac{1}{4} ⇒λ = ± \frac{1}{2}⇒[(i)] y = ±x$.
5. Substitute back into the constraint: x²+(±x)²=-1 ↭ 2x²=1 ⇒ $x=±\frac{1}{\sqrt{2}}$, therefore there are four possibilities y = x, $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$, y = -x, $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$.
6. Finally, we evaluate f at these critical points: $f(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}+1 = \frac{1}{2}+1 = \frac{3}{2}, f(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})=(-\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})+1 = \frac{1}{2}+1 = \frac{3}{2}, f(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})=(\frac{1}{\sqrt{2}})(-\frac{1}{\sqrt{2}})+1 = -\frac{1}{2}+1 = \frac{1}{2}, f(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})=(-\frac{1}{\sqrt{2}})(\frac{1}{\sqrt{2}})+1 = -\frac{1}{2}+1 = \frac{1}{2}$.
7. Conclusion. The maximum value of f subject to the constraint g is $\frac{3}{2}$, and it occurs at the points $(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$. The minimum value of f subject to the constraint g is $\frac{1}{2}$, and it occurs at the points $(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}), (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ (Figures A and B).

• Calculate maximum/minimum of f(x, y, z) = 3x²+y²-2z², with the constraint: g(x, y) = 3x+2y-8z = -50

In this problem, we are tasked with finding the maximum and minimum values of the function f(x, y, z) = 3x²+y²-2z² under the constraint g(x, y) = 3x +2y -8z = -50. The constraint represents a plane in 3D space. To solve this problem, we will use the method of Lagrange multipliers, which helps find extrema of functions subject to constraints.

1. Formulate the System Using Lagrange Multipliers. The method of Lagrange multipliers involves introducing a new variable, λ (the Lagrange multiplier), and asserts that at an extremum of f(x, y, z) subject to the constraint g(x, y, z) = 0, the gradients of f and g must be parallel: ∇f = λ∇g. This results in the following system of equations: $\begin{cases} f_x = λg_x \\ f_y = λg_y \\ f_z = λg_z \\ g(x, y, z) = 0\end{cases}$.
2. Compute the Gradients. Let’s compute the partial derivatives (gradients) of f(x, y, z) and g(x, y, z). ∇f(x, y, z) = ⟨6x,2y,−4z⟩, ∇g(x, y, z) = ⟨3, 2, −8⟩.
3. Set Up the System of Equations. Using the gradients calculated above, we set up the system of equations based on the relationship ∇f = λ∇g: $\begin{cases} 6x = 3λ (i) \\ 2y = 2λ (ii) \\ -4z = -8λ (iii) \\ g = 0\end{cases}$.
4. Solve the system of equations: From equation (i)⇒$x=\frac{1}{2}λ.$ From equation (ii)⇒y = λ. From equation (iii)⇒$z = 2λ.$
5. Substitute back into the constraint: g(x, y) = 3x +2y -8z = -50 ⇒ $3\frac{1}{2}λ+2λ-8(2λ) = -50⇒ \frac{3}{2}λ-14λ = -50⇒[\text{Multiplying by 2}] 3λ -28λ = -100 ⇒[\text{This simplifies to}] -25λ = -100⇒λ = \frac{-100}{-25}=4$.
6. Determine the values of x, y, and z. Substituting λ = 4 back into our expressions for x, y, and z:, $x =\frac{1}{2}λ = \frac{1}{2}·4 = 2, y = λ = 4, z = 2λ = 2·4 = 8.$
7. Evaluate f at the critical point ⟨2, 4, 8⟩. $f(2, 4, 8) = 3·2^2+4^2-2·8^2 = 3·4+16-2·64 = 12 + 16 -128 = -100.$
8. Conclusion. The critical point found is (2, 4, 8) and the value of the function f(x, y, z) at this point is −100. Therefore, the critical point (2, 4, 8) gives the minimum value of f(x, y, z) subject to the constraint g(x,y,z) = 0.

To confirm whether this point represents a minimum, it can be checked that all other points that satisfies the constrain g(x, y, z) = 3x +2y-8z = -50, e.g., (4, -3, 7), (0, 11, 9), their values is less than -100 (f(4, -3, 7) = f(0, 11, 9) = -41 > -100)

• Building a Pyramid with a Given Triangular Base and Volume

We are tasked with constructing a pyramid that has a specific triangular base and a predetermined volume. Our objective is to minimize the total surface area of the pyramid while maintaining these constraints. (Figure i, ii).

1. Understanding the problem. The volume V of pyramid with a triangular base is given by ¹⁄₃Area_base·h and both Area_base, the area of the triangular base, and h, the height of the pyramid are fixed or known, hence the height h can be determined and expressed in terms of the volume and the base area h = $\frac{V}{Area_{base}}$

2. Surface Area of the Pyramid. The total surface area of the pyramid consists of the area of the triangular base (which is fixed) and the areas of the three triangular faces (sides) that connect the apex to each edge of the base. Let u₁, u₂, and u₃ be the distances from Q to the sides of the triangular base (P₁, P₂, and P₃) where Q(x, y, 0) is the point on the base of the pyramid and P(x, y, h) is the peak.

   We need to minimize the total surface area of the pyramid. The surface area consists of the areas of the three triangular sides (the area of the base is fixed). The area of each triangular side can be expressed as: $A_{side, i} = \frac{1}{2}a_i\sqrt{u_i^2+h^2}$ where a_i is the length of the i-th side of the base and $\sqrt{u_i^2+h^2}$ is the slant height of the triangular face. Notice that there is a right triangle with cathetus u_i and h, and the hypotenuse is $\sqrt{u_i^2+h^2}$.

   Therefore, the total side surface area is $f(u_1, u_2, u_3) = \frac{1}{2}a_1\sqrt{u_1^2+h^2}+\frac{1}{2}a_2\sqrt{u_2^2+h^2}+\frac{1}{2}a_3\sqrt{u_3^2+h^2} = $.

3. Setting up the Optimization Problem. The constraint is the fixed area of the triangular base, Area_base = g(u₁, u₂, u₃) = constant. Each of these sides have base, height: $(a_1,u_1), (a_2,u_2),~ and~ (a_3,u_3)$ respectively, so the total Area_base = $\frac{1}{2}a_1u_1 + \frac{1}{2}a_2u_2 + \frac{1}{2}a_3u_3$

4. Applying Lagrange Multipliers To minimize the surface area subject to the constraint on the base area, we apply the method of Lagrange multipliers. This method involves finding the points where the gradients of the surface area function f(u₁, u₂, u₃) = S and the constraint function g(u₁, u₂, u₃) are parallel. Mathematically, this condition is expressed as: ∇f = λ ∇fg.

5. Compute the Gradients. Let’s compute the partial derivatives (gradients) of f and g. $\frac{∂f}{u_1} = \frac{1}{2}a_1\frac{u_1}{\sqrt{u_1^2+h^2}}, \frac{∂f}{u_2} = \frac{1}{2}a_2\frac{u_2}{\sqrt{u_2^2+h^2}}, \frac{∂f}{u_3} = \frac{1}{2}a_3\frac{u_3}{\sqrt{u_3^2+h^2}}$. $\frac{∂g}{u_1} = \frac{1}{2}a_1, \frac{∂g}{u_2} = \frac{1}{2}a_2, \frac{∂g}{u_3} = \frac{1}{2}a_3$

6. Setting the System of Equations. Using the gradients calculated above, we set up the system of equations based on the relationship ∇f = λ∇g: $\frac{∂f}{u_1} = \frac{1}{2}a_1\frac{u_1}{\sqrt{u_1^2+h^2}} = λ\frac{1}{2}a_1 ↭ \frac{u_1}{\sqrt{u_1^2+h^2}} = λ,$ and similarly, $\frac{u_2}{\sqrt{u_2^2+h^2}} = λ, \frac{u_3}{\sqrt{u_3^2+h^2}} = λ ⇒ \frac{u_1}{\sqrt{u_1^2+h^2}} = \frac{u_2}{\sqrt{u_2^2+h^2}} = \frac{u_3}{\sqrt{u_3^2+h^2}}$

7. Solve the system of equations: $λ = \frac{u_1}{\sqrt{u_1^2+h^2}} = \frac{u_2}{\sqrt{u_2^2+h^2}} = \frac{u_3}{\sqrt{u_3^2+h^2}}$. Since the distance u₁, u₂, and u₃ are proportional to each other, and all sides of the pyramid are symmetric in this configuration, it follows that: u₁ = u₂ = u₃⇒The solution implies that Q is located at the incenter of the triangular base.

Recall that the incenter is the point where the three interior angle bisectors of a triangle meet. The incenter is equidistant from the sides of the triangle.

• Find the dimensions of a closed cylindrical field jar with volume 2000 cm³ that yield a minimum surface area.

We are given a closed cylindrical jar with a fixed volume of 2000 cm³. The objective is to find the dimensions (radius r and height h) that will minimize the surface area of the jar.

1. Understanding the problem. The jar volume (the volume of a cylinder) is given by g(r, h) = πr²h = 2000 where r is the radius of the base, and h is the height of the cylinder. The surface area of the cylinder is f(r, h) = 2πr²h + 2πrh where 2πr² represents the combined areas of the top and bottom circles, and 2πrh represents the lateral surface area (which is the area of the rectangle formed by unrolling the side of the cylinder, with base 2πr and height h).
2. Formulate the System Using Lagrange Multipliers. To minimize the surface area A(r, h) subject to the volume constraint V(r,h), we can use the method of Lagrange multipliers. It involves introducing a new variable, λ (the Lagrange multiplier), and asserts that at an extremum of A(r, h) subject to the constraint V(r, h) = 2000, the gradients of A and V must be parallel: ∇A = λ∇V.
3. Compute the Gradients. Let’s compute the partial derivatives (gradients) of A(r, h) and V(r, h). ∇A(r, h) = ⟨4πr + 2πh, 2πr⟩, ∇V(r, h) = ⟨2πrh, πr²⟩.
4. Setting the System of Equations. Using the gradients calculated above, we set up the system of equations based on the relationship ∇f = λ∇g: (i) 4πr + 2πh = 2πrhλ, (2) 2πr = πr²λ, and the volume constraint (3) πr² = 2000.
5. Solve the system of equations. From (i): 2r + h = rhλ. Solve for λ from Equation 2: 2 = rλ ↭ λ = $\frac{2}{r} ⇒[\text{Substitute λ into (i)}] 4πr + 2πh = 2πrh·\frac{2}{r} ↭ 2r + 2h = 4h ↭ h = 2r$.
6. Substitute back into the constraint: πr²h = 2000 ↭ πr²·2r = 2000 ↭ $r^2 = \frac{1000}{π} ⇒ r = \frac{10}{\sqrt[3]{π}}cm, h = \frac{20}{\sqrt[3]{π}}cm$. These dimensions will minimize the surface area of the cylindrical jar while maintaining a volume of 2000 cm³. This result implies that the optimal cylindrical jar will have a height that is twice its radius.

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.