lagrange multiplier example problems

and the equality constraints are affine functions which have the general form: As a side note, affine functions are convex. The curve $100=2x+2y$ can be thought of as a level curve of the Use the problem-solving strategy for the method of Lagrange multipliers with an objective function of three variables. Therefore, the system of equations that needs to be solved is, \[\begin{align*} 2 x_0 - 2 &= \lambda \\ 8 y_0 + 8 &= 2 \lambda \\ x_0 + 2 y_0 - 7 &= 0. Let’s have another observation regarding the inequality constraint. At some point you will see the g is a convex function because it is an affine function. The feasible set for x in a linear program is a polyhedron. In that example, the constraints involved a maximum number of golf balls that could be produced and sold in $1$ month $(x),$ and a maximum number of advertising hours that could be purchased per month $(y)$. which $\ds f(x,y,z) = xy^2z^2$ has a maximum value. If it is true, x*, α* and λ* are the primal and dual solution or vice versa if strong duality holds. The plane $x-y+z=2$ intersects the cylinder $x^2+y^2=4$ in an Set up a system of equations using the following template: \[\begin{align} \vecs ∇f(x_0,y_0) &=λ\vecs ∇g(x_0,y_0) \\[4pt] g(x_0,y_0) &=0 \end{align}.\]. How

to the constraint $\ds g(x,y,z) = 4x^2+2y^2 + z^2 - 70 = 0$. following system to solve:

Neither of these values exceed $540$, so it seems that our extremum is a maximum value of $f$, subject to the given constraint. The goal is still to maximize profit, but now there is a different type of constraint on the values of $x$ and $y$. The property that p* = d* is called the strong duality. the equation $A=xy$ defines a surface, and the equation $100=2x+2y$ Since the point $(x_0,y_0)$ corresponds to $s=0$, it follows from this equation that, \[\vecs ∇f(x_0,y_0)⋅\vecs{\mathbf T}(0)=0, \nonumber\], which implies that the gradient is either the zero vector $\vecs 0$ or it is normal to the constraint curve at a constrained relative extremum. Then there is a number $λ$ called a Lagrange multiplier, for which, \[\vecs ∇f(x_0,y_0)=λ\vecs ∇g(x_0,y_0).\], Assume that a constrained extremum occurs at the point $(x_0,y_0).$ Furthermore, we assume that the equation $g(x,y)=0$ can be smoothly parameterized as.
$f(2,1,2)=9$ is a minimum value of $f$, subject to the given constraints. convenient, the method of Lagrange multipliers. Recall that the gradient of a function of more than one variable is a vector. figure 14.8.3.

\end{align*}\] Therefore, either $z_0=0$ or $y_0=x_0$. \end{align*}\]. the vectors, but now in five unknowns, $x$, $y$, $z$, $\lambda$, and This property is called the weak duality.

Intuitively, we break down an objective function into smaller sub-problems. So let’s view it from a different perspective and discover the relationship between the following minmax solution and the maxmin solution as a possible alternative. Use the problem-solving strategy for the method of Lagrange multipliers. $$\nabla f =\langle 2x,2y,2z\rangle\qquad $h(x,y,z)=c_2$. \sqrt{9-x^2-y^2}$ when $\ds x^2+y^2 \leq 9$. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. \end{align*}\], The first three equations contain the variable $λ_2$. In other words, we want to have a line with the given slope -α that has the smallest y-intersection providing that it still in contact with the green zone. 2y&=\lambda 2y+\mu\cr In a convex set, any value between two points in the set must belong to the convex set also. For example, in a future course or courses in Physics (e.g., thermal physics, statistical mechanics), you should see a derivation of the famous “Boltzman distribution” of the energies of atoms in an ideal gas using Lagrange multipliers. provides a fourth equation. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. This constraint and the corresponding profit function, \[f(x,y)=48x+96y−x^2−2xy−9y^2 \nonumber\]. interest, and the origin. We then substitute $(10,4)$ into $f(x,y)=48x+96y−x^2−2xy−9y^2,$ which gives \[\begin{align*} f(10,4) &=48(10)+96(4)−(10)^2−2(10)(4)−9(4)^2 \\[4pt] &=480+384−100−80−144 \\[4pt] &=540.\end{align*}\] Therefore the maximum profit that can be attained, subject to budgetary constraints, is $$540,000$ with a production level of $10,000$ golf balls and $4$ hours of advertising bought per month. In the middle figure, we add an inequality constraint to the optimization problem. The Lagrangian is L = f(x) + α l(x). Double Integrals in Cylindrical Coordinates, 3. We start with a random guess of λ and use any optimization method to solve the unconstrained objective. optimized is a function of three variables and the constraint represents So J is a perfect cost function for our optimization problem that enforcing the inequality constraint. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. \end{align*}\], We use the left-hand side of the second equation to replace $λ$ in the first equation: \[\begin{align*} 48−2x_0−2y_0 &=5(96−2x_0−18y_0) \\[4pt]48−2x_0−2y_0 &=480−10x_0−90y_0 \\[4pt] 8x_0 &=432−88y_0 \\[4pt] x_0 &=54−11y_0.
the origin. Subtracting the first two we get that can be sent in a rectangular box? Otherwise, J is simply f(x). 2x&=\lambda 2x+\mu\cr function, and we look for points at which the two gradients are \] Recall $y_0=x_0$, so this solves for $y_0$ as well. In ML, we often transform, approximate or relax our problems into one of these easier optimization models. Find the points on the ellipse closest to and farthest from Find $x$ and $\phi$ so that the Find three real numbers whose sum is 9 and the sum of whose squares is an example of an optimization problem, and the function $f(x,y)$ is called the objective function. The solution we already understand effectively The objective function is $f(x,y,z)=x^2+y^2+z^2.$ To determine the constraint functions, we first subtract $z^2$ from both sides of the first constraint, which gives $x^2+y^2−z^2=0$, so $g(x,y,z)=x^2+y^2−z^2$.

Next, we calculate $\vecs ∇f(x,y,z)$ and $\vecs ∇g(x,y,z):$ \[\begin{align*} \vecs ∇f(x,y,z) &=⟨2x,2y,2z⟩ \\[4pt] \vecs ∇g(x,y,z) &=⟨1,1,1⟩. $\langle 2,2\rangle=\lambda\langle y,x\rangle$, that is, when Our soluting will be one of the points in the green area. value. Section 6.4 – Method of Lagrange Multipliers 241 Example 2: Find the minimum and maximum points on the surface f (x, y) =x2 +y2 −2x −2y subject to the constraint x2 +y2 =4. The rest will be simple because the result function g will be convex and easy to optimize. The figure shows the cylinder, the plane, the four points of

Since each of the first three equations has $λ$ on the right-hand side, we know that $2x_0=2y_0=2z_0$ and all three variables are equal to each other.

Use the problem-solving strategy for the method of Lagrange multipliers with two constraints.

Use the problem-solving strategy for the method of Lagrange multipliers with an objective function of three variables. 2z&=0-\mu\cr \end{align*}\] This leads to the equations \[\begin{align*} ⟨2x_0,2y_0,2z_0⟩ &=λ⟨1,1,1⟩ \\[4pt] x_0+y_0+z_0−1 &=0 \end{align*}\] which can be rewritten in the following form: \[\begin{align*} 2x_0 &=λ\\[4pt] 2y_0 &=λ \\[4pt] 2z_0 &=λ \\[4pt] x_0+y_0+z_0−1 &=0. curve have the same slope—their tangent lines are parallel. Let’s follow the problem-solving strategy: 1. We then substitute this into the first equation, \[\begin{align*} z_0^2 &= 2x_0^2 \\[4pt] (2x_0^2 +1)^2 &= 2x_0^2 \\[4pt] 4x_0^2 + 4x_0 +1 &= 2x_0^2 \\[4pt] 2x_0^2 +4x_0 +1 &=0, \end{align*}\] and use the quadratic formula to solve for $x_0$: \[ x_0 = \dfrac{-4 \pm \sqrt{4^2 -4(2)(1)} }{2(2)} = \dfrac{-4\pm \sqrt{8}}{4} = \dfrac{-4 \pm 2\sqrt{2}}{4} = -1 \pm \dfrac{\sqrt{2}}{2}. The largest of the values of $f$ at the solutions found in step $3$ maximizes $f$; the smallest of those values minimizes $f$.

We write down }$$ Or we can check which value of yᵢ in the later iterations will yield the most optimal solution. \nonumber\]To ensure this corresponds to a minimum value on the constraint function, let’s try some other points on the constraint from either side of the point $(5,1)$, such as the intercepts of $g(x,y)=0$, Which are $(7,0)$ and $(0,3.5)$. 100 units. the package perpendicular to the length). From another perspective, in a convex optimization problem, f and l are convex functions. parallel, giving us three equations in four unknowns. the origin. \end{align*}\] The two equations that arise from the constraints are $z_0^2=x_0^2+y_0^2$ and $x_0+y_0−z_0+1=0$. Once, we have the constraints added, we solve it with the Lagrangian multiplier.