Lagrange Multipliers

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

Functions, Partials, and Gradients

• A function of two variables $f: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ assigns to each ordered pair in its domain a unique real number. Examples include geometric formulas like Area = $\frac{1}{2}bh$, or algebraic surfaces like $z = f(x, y) = 2x + 3y$, $f(x, y) = x^2 + y^2$, and $f(x,y) = e^{x+y}$.

• Partial derivatives differentiate a multivariable function $f(x_1, x_2, \dots, x_n)$ with respect to one target variable while treating all other variables as constants. They measure how the function changes in a single specific direction.
  If $f(x, y) = 2x^2y^3$, then $\frac{\partial f}{\partial x} = f_x = 4xy^3$ and $\frac{\partial f}{\partial y} = f_y = 6x^2y^2$.
  If $f(x, y) = x^3y + y^2$, then $f_x = 3x^2y$ and $f_y = x^3+2y$.

• A point $(x_1, x_2, \dots, x_n)$ is considered a critical point if the gradient of $f$ at that point is the zero vector, or if the gradient does not exist.

Gradients, Level Surfaces, and Directional Derivatives

• The gradient $\nabla f$ is a vector containing all first-order partial derivatives of $f$. It geometrically points in the direction of the function’s steepest ascent. $\nabla f = \left\langle \frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \dots, \frac{\partial f}{\partial x_n} \right\rangle$.

• In 2D space ($n = 2$), finding a critical point $(x_0, y_0)$ requires solving the system: $\frac{\partial f}{\partial x}(x_0, y_0) = 0$ and $\frac{\partial f}{\partial y}(x_0, y_0) = 0$.

• Theorem: The gradient vector is perpendicular (orthogonal) to the level curves or level surfaces of the function at any given point. This means if you walk along a contour line where the function’s value is constant, the direction of steepest uphill is exactly 90 degrees to your path.

The directional derivative $\frac{dw}{ds}\bigg\vert{}_{\vec{u}}$ represents the rate of change of a function $w$ as you move in the direction of a unit vector $\vec{u}$. It can be expressed or compactly written via the dot product of the gradient and the direction vector: $\frac{dw}{ds}\bigg\vert{}_{\vec{u}} = \nabla w \cdot \vec{u} = \vert{}\nabla w\vert{}\vert{}\vec{u}\vert{}\cos(\theta) = \vert{}\nabla w\vert{}\cos(\theta)$ where $\theta$ is the angle between the gradient $\nabla w$ and the unit vector $\vec{u}$.

Lagrange Multipliers

Lagrange multipliers are a powerful method for finding the local maxima and minima of a function, $f(x, y, z)$, where the variables $x, y$, and $z$ are not entirely independent, but are instead constrained by an equation, $g(x, y, z) = c$.

1. When you look for extrema of $f(x, y, z)$ subject to a constraint ($g(x, y, z) = c$), you are no longer free to move in all directions. You can only move along the constraint surface.

2. 🌄 The equation $f(x, y, z) = k$ defines a level surface of $f$. The equation $g(x, y, z) = c$ defines the constraint surface on which you must stay. If you imagine sliding along the constraint surface, you are looking for places where $f$ is as large or as small as possible while staying on that surface. Therefore, the level surface of $f$ that you “touch” must be tangent to the constraint surface.

3. If the two surfaces are tangent, their gradient vectors, which are perpendicular to these level curves, must be parallel. This relationship can be expressed mathematically as: $\nabla f = \lambda \nabla g$ where $\lambda$ is the Lagrange multiplier. The gradients $\nabla f$ and $\nabla g$ represent the directions of the steepest ascent for the functions f and g, respectively.

4. 🎯 Gradients are perpendicular to level surfaces. $\nabla f$ is perpendicular to the level surface $f = k$. $\nabla g$ is perpendicular to the constraint surface $g = c$.
   

   A constrained extremum is a point where you cannot move along the constraint surface in any direction that changes 𝑓. That single fact forces the gradient of 𝑓 to be perpendicular to every tangent direction of the constraint surface.

   Imagine you are hiking along a narrow path carved into a mountain defined by the constraint line $g = c$. As you walk, you track the altitude $f$. You are not free to wander anywhere on the mountain — you must stay on the path. So the only directions you can move are tangent to the path. If the altitude is still rising or falling as you step an infinitesimal step forward, you have not reached the peak of your path. You only reach the highest (or lowest) point on your path when the altitude stops changing for a brief moment (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).
   At this exact peak, the directional derivative (rate of change) in your direction of travel $\vec{a}$ is exactly zero, $\frac{df}{ds}\big\vert{}_{\vec{a}} = \nabla f \cdot \vec{a} = 0$ (Figure E).
   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.
   If $\nabla f \cdot \vec{a} = 0$, then $\nabla f$ is perpendicular to $\vec{a}$ (and thus perpendicular to the constraint surface). Since $\nabla g$ is also defined as being perpendicular to the surface $g = c$, the two gradients must point in the same (or opposite) direction: $\nabla f \parallel \nabla g$.

Note on Classification: The method of Lagrange multipliers yields critical points but does not inherently classify them as maxima, minima, or saddle points. You must evaluate $f(x,y,z)$ at each critical point to determine which is the absolute maximum or minimum under the constraint.

To solve these problems, we set up a system of equations:

$$\begin{cases} f_x = \lambda g_x \\ f_y = \lambda g_y \\ f_z = \lambda g_z \\ g = c \end{cases}$$

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. Therefore, Objective Function: $f(x, y) = x^2 + y^2$ (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, hill or parabolic bowl subject to the fact you can only walk along a specific trail projected onto it (g).

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

$\begin{cases} \frac{\partial f}{\partial x} = \lambda \frac{\partial g}{\partial x} \\ \frac{\partial f}{\partial y} = \lambda \frac{\partial g}{\partial y} \\ g = c \end{cases}$

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.

$\begin{cases} f_x = \lambda g_x \\ f_y = \lambda g_y \\ g = c\end{cases}$

In our particular example, $f(x, y) = x^2 + y^2, g = xy$. Compute Gradients: $\nabla f = \langle 2x, 2y \rangle$, $\nabla g = \langle y, x \rangle$.

$$\begin{cases} 2x = \lambda y \quad \text{(i)} \\ 2y = \lambda x \quad \text{(ii)} \\ xy = 3 \quad \text{(iii)} \end{cases}$$

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

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

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

1. Case A $\lambda = 2 \implies 2x -2y = 0 ⇒ x = y, x^2 = 3$. There are two solutions: $(\sqrt{3},\sqrt{3})$ or $(-\sqrt{3},-\sqrt{3})$.
2. Case B $\lambda = -2 \implies 2x +2y = 0 \implies x = -y, -x^2 = 3$, which has no real 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, $\sqrt{3+3}=\sqrt{6}$

• Find the dimensions of the largest rectangular box in the first octant with three faces on the coordinate 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, $\lambda$ (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: $\nabla f \parallel \nabla g$. This gives us the following system of equations: $\begin{cases} f_x = \lambda g_x \\ f_y = \lambda g_y \\ g = 0\end{cases}$
2. Compute the Gradients. Let’s compute the partial derivatives f and g: $\nabla f = \langle yz, xz, xy \rangle$, $\nabla g = \langle 2, 3, 5 \rangle$.
3. Set up the equations. Using the gradients calculated above, we set up the system of equations based on the relationship $\nabla f \parallel \nabla g$: $\begin{cases} yz = 2\lambda \quad \text{(i)} \\ xz = 3\lambda \quad \text{(ii)} \\ xy = 5\lambda \quad \text{(iii)} \\ 2x + 3y + 5z = 90 \quad \text{(iv)} \end{cases}$
4. Solve the system of equations. From equation (i), $\lambda = \frac{yz}{2}$. From equation (ii), $\lambda = \frac{xz}{3}$. From equation (iii), $\lambda = \frac{xy}{5}$. As we want to maximize the volume, we do not need to consider z = 0 or y = 0. Equate the first and second expressions: $\frac{yz}{2} = \frac{xz}{3} \implies \frac{y}{2} = \frac{x}{3} \implies y = \frac{2}{3}x$. Equate the first and third expressions: $\frac{yz}{2} = \frac{xy}{5} \implies \frac{z}{2} = \frac{x}{5} \implies z = \frac{2}{5}x$
5. Substitute back into the constraint: $2x + 3y + 5z = 90 \implies 2x +3(\frac{2}{3}x) + 5(\frac{2}{5}x) = 90 \iff [\text{Simplify the equation:}] 2x + 2x +2x = 90 \iff 6x = 90 \iff x = \frac{90}{6} = 15$.
   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 $f(x, y, z)= xyz$: $f(15, 10, 6) = 15 \cdot 10 \cdot 6 = 900 \text{ cubic units}$

• Calculate maximum/minimum of $f(x, y) = xy+1$ subject to the constraint $x^2 + y^2 = 1$.

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 $g(x, y) = x^2 + y^2 - 1$. 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, $\lambda$ (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: $\nabla f \parallel \nabla g$. This gives us the following system of equations: $\begin{cases} f_x = \lambda g_x \\ f_y = \lambda g_y \\ g = 0\end{cases}$
2. Compute the Gradients. Let’s compute the partial derivatives f and g: $\nabla f = \langle y, x \rangle$, $\nabla g = \langle 2x, 2y \rangle$.
3. Set up the equations. Using the gradients calculated above, we set up the system of equations based on the relationship $\nabla f \parallel \nabla g$: $\begin{cases} y = 2\lambda x \quad \text{(i)} \\ x = 2\lambda y \quad \text{(ii)} \\ x^2 + y^2 = 1 \quad \text{(iii)} \end{cases}$
4. Solve the system of equations. From equation (i), $y = \lambda 2x$. From equation (ii), $x = 2\lambda y$. We can substitute x from Equation (ii) into Equation (i): $y = \lambda 2(\lambda 2y)=[(ii)]4\lambda^2y$. There are two solutions: $y = 0 \implies [\text{By (ii)}] x = 0$, but $(0, 0)$ does not satisfy our constrain $0^2 + 0^2 -1 = -1 \ne 0$ or $y \ne 0 \implies[\text{Dividing by y}] 1 = 4 \lambda^2 \implies \lambda^2 = \frac{1}{4} \implies \lambda = \pm \frac{1}{2}⇒[(i)] y = \pm x$.
5. Substitute back into the constraint: $x^2 + (\pm x)^2 = 1 \iff 2x^2 =1 \implies x = \pm \frac{1}{\sqrt{2}}$, therefore there are four possibilities, namely (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 $f(x, y, z) = 3x^2 +y^2 -2z^2$, with the constraint: $g(x, y, z) = 3x+2y-8z = -50$

In this problem, we are tasked with finding the maximum and minimum values of the function $f(x, y, z) = 3x^2 +y^2 -2z^2$ 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, $\lambda$ (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: $\nabla f \parallel \nabla g$. This results in the following system of equations: $\begin{cases} f_x = \lambda g_x \\ f_y = \lambda g_y \\ f_z = \lambda g_z \\ g(x, y, z) = 0\end{cases}$.
2. Compute the Gradients. Let’s compute the partial derivatives (gradients) of $f(x, y, z)$ and $g(x, y, z)$. $\nabla f(x, y, z) = \langle 6x, 2y, -4z \rangle$, $\nabla g(x, y, z) = \langle 3, 2, -8 \rangle$.
3. Set Up the System of Equations. Using the gradients calculated above, we set up the system of equations based on the relationship $\nabla f \parallel \nabla g$: $\begin{cases} 6x = 3 \lambda \quad \text{(i)} \\ 2y = 2\lambda \quad \text{(ii)} \\\ -4z = -8\lambda \quad \text{(iii)} \\\ g = 0 \quad \text{(iv)}\end{cases}$.
4. Solve the system of equations: From equation (i), $x=\frac{1}{2}\lambda.$ From equation (ii), $y = \lambda$. From equation (iii), $z = 2\lambda.$
5. Substitute the expressions for $x$, $y$, and $z$ back into the constraint: $g(x, y) = 3x +2y -8z = -50$. $3\left(\frac{1}{2}\lambda\right) + 2(\lambda) - 8(2\lambda) = -50 \implies \frac{3}{2}\lambda + 2\lambda - 16\lambda = -50$. Multiplying by 2 to clear the fraction: $3\lambda + 4\lambda - 32\lambda = -100 \implies[\text{This simplifies to}] -25\lambda = -100 \implies \lambda = 4$.
6. Determine the values of x, y, and z. Substituting $\lambda = 4$ back into our expressions for x, y, and z:, $x =\frac{1}{2}\lambda = \frac{1}{2}\cdot 4 = 2, y = \lambda = 4, z = 2\lambda = 2\cdot 4 = 8.$
   The critical point is (2, 4, 8).
7. Evaluate f at the critical point (2, 4, 8). $f(2, 4, 8) = 3\cdot 2^2+4^2-2·8^2 = 3\cdot 4+16-2·64 = 12 + 16 -128 = -100.$
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 $f(x, y, z) = 3x^2 +y^2 -2z^2$ subject to the constraint $g(x, y) = 3x+2y-8z = -50$.

To confirm whether this point represents a minimum, it can be checked that all other points that satisfies the constrain $g(x, y) = 3x+2y-8z = -50$, e.g., (4, -3, 7), (0, 11, 9), their values are greater 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 ($A_{\text{base}}$) and a predetermined volume ($V$). 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 $V = \frac{1}{3} A_{\text{base}} h$. Since $V$ and $A_{\text{base}}$ are fixed, the height $h$ is also constant: $h = \frac{3V}{A_{\text{base}}}$.
   Let the base be a triangle with side lengths $a_1, a_2, a_3$. Let point $Q(x, y, 0)$ be the projection of the pyramid’s peak onto the base, and $u_1, u_2, u_3$ be the perpendicular distances from $Q$ to each side of the 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.
   The area of the base is fixed, so we only minimize the lateral surface area ($S$). Each triangular face has a base $a_i$ and a slant height $\sqrt{u_i^2 + h^2}$ (forming a right triangle with the height $h$ and base distance $u_i$). 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, $A_{\text{base}} = g(u_1, u_2, u_3) = constant$. This is the sum of three smaller triangles formed by connecting $Q$ to the vertices of the base. Each of these triangles have base, height: $(a_1,u_1), (a_2,u_2),~ and~ (a_3,u_3)$ respectively, so the total $A_{\text{base}} = \frac{1}{2}a_1u_1 + \frac{1}{2}a_2u_2 + \frac{1}{2}a_3u_3$
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 $f(u_1, u_2, u_3)$ and the constraint function $A_{\text{base}} = \frac{1}{2}a_1u_1 + \frac{1}{2}a_2u_2 + \frac{1}{2}a_3u_3$ are parallel. Mathematically, this condition is expressed as: $\nabla f = \lambda \nabla g$.
5. Compute the Gradients. Let’s compute the partial derivatives (gradients) of f and g. $\frac{\partial f}{\partial u_1} = \frac{1}{2}a_1\frac{u_1}{\sqrt{u_1^2+h^2}}, \frac{\partial f}{\partial u_2} = \frac{1}{2}a_2\frac{u_2}{\sqrt{u_2^2+h^2}}, \frac{\partial f}{\partial u_3} = \frac{1}{2}a_3\frac{u_3}{\sqrt{u_3^2+h^2}}$. $\frac{\partial g}{\partial u_1} = \frac{1}{2}a_1, \frac{\partial g}{\partial u_2} = \frac{1}{2}a_2, \frac{\partial g}{\partial u_3} = \frac{1}{2}a_3$
6. Setting the System of Equations. Using the gradients calculated above, we set up the system of equations based on the relationship $\nabla f \parallel \nabla g$: $\frac{\partial f}{\partial u_1} = \frac{1}{2}a_1\frac{u_1}{\sqrt{u_1^2+h^2}} = \lambda \frac{1}{2}a_1 \implies \frac{u_1}{\sqrt{u_1^2+h^2}} = \lambda,$ and similarly, $\frac{u_2}{\sqrt{u_2^2+h^2}} = \lambda, \frac{u_3}{\sqrt{u_3^2+h^2}} = \lambda \implies \frac{u_1}{\sqrt{u_1^2+h^2}} = \frac{u_2}{\sqrt{u_2^2+h^2}} = \frac{u_3}{\sqrt{u_3^2+h^2}}$
7. Solve the system of equations: $\lambda = \frac{u_1}{\sqrt{u_1^2+h^2}} = \frac{u_2}{\sqrt{u_2^2+h^2}} = \frac{u_3}{\sqrt{u_3^2+h^2}}$.
   Since $h$ is constant for all faces (every single lateral face drops down from that exact same height h to the base plane), $k(u) = \frac{u}{\sqrt{u^2 + h^2}}$ is strictly increasing (one-to-one function) for $u \gt 0$, this equality is only true if $u_1 = u_2 = u_3$.
   Key insight: The peak of the pyramid must be placed directly above the point that is equidistant from all three sides of the triangular base. In geometry, this point is exactly the incenter of the triangle.

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 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 $V(r, h) = \pi r^2 h = 2000$ where r is the radius of the base, and h is the height of the cylinder.
   The surface area of the cylinder is $A(r, h) = 2\pi r^2 + 2\pi rh$ where $2\pi r^2$ is the area of the top and bottom circles, and $2\pi rh$ is the lateral area (which is the area of the rectangle formed by unrolling the side of the cylinder, with base $2\pi r$ 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, $\lambda$ (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: $\nabla A = \lambda \nabla V$.
3. Compute the Gradients. Let’s compute the partial derivatives (gradients) of A(r, h) and V(r, h): $\nabla A(r, h) = \langle 4\pi r + 2\pi h, 2\pi r \rangle$ and $\nabla V(r, h) = \langle 2\pi rh, \pi r^2 \rangle$
4. Setting the System of Equations. Using the gradients calculated above, we set up the system of equations based on the relationship $\nabla f \parallel \nabla g$: (i) $4\pi r + 2\pi h = \lambda(2\pi rh)$, (ii) $2\pi r = \lambda(\pi r^2)$, and (iii) $\pi r^2 h = 2000$.
5. Solve the system of equations. From equation (ii), divide by $\pi r$ (assuming $r \neq 0$): $2 = \lambda r \implies \lambda = \frac{2}{r}$.
   Substitute $\lambda$ into equation (i): $4\pi r + 2\pi h = \left(\frac{2}{r}\right)(2\pi rh) \implies 4\pi r + 2\pi h = 4\pi h$
   Divide by $2\pi$: $2r + h = 2h \implies h = 2r$
6. Substitute $h = 2r$ back into the volume constraint: $\pi r^2(2r) = 2000 \implies 2\pi r^3 = 2000 \implies r^3 = \frac{1000}{\pi}$.
   Therefore, $r = \frac{10}{\sqrt[3]{\pi}} \text{ cm}$ and $h = 2r = \frac{20}{\sqrt[3]{\pi}} \text{ cm}$
7. Conclusion: 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 exactly twice its radius or equal to its diameter.

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.