Name: optimizing a fence pen with lagrange multipliers
Uploaded: 2026-06-08T09:38:27.653Z
Duration: 1 min 29 s
Channel: JoVE

A rectangular pen is to be built against a barn wall using 100 meters of fencing. Because the wall forms one side, only three sides need fencing. The goal is to find the dimensions that give the largest area.This is a constrained optimization problem because the area must be maximized while the fencing length remains fixed.Lagrange multipliers give a clear way to include this restriction. In the length-width plane, the fencing condition forms a constraint curve that limits the possible pen dimensions. The maximum happens where a level curve of the area touches the constraint curve. At that point, both curves share a common tangent, so their gradients are parallel.The gradients may have different sizes, so one is multiplied by a constant to match the other. This constant is the Lagrange multiplier, lambda.The next step is to take partial derivatives with respect to length and width. Solving these equations shows that the length is twice the width. Substituting this relationship into the fencing constraint gives the two dimensions. These dimensions give the largest possible area for the rectangular pen. This shows how Lagrange multipliers connect optimization, geometry, and constraints.

lagrange multipliers: one constraint

In constrained optimization, the objective is to maximize or minimize a quantity while satisfying a fixed condition. A standard example is a rectangular pen built against a barn wall using 100 meters of fencing. Because the wall provides one side of the enclosure, only the other three sides require fencing. The problem is to find the dimensions that produce the greatest possible area.Let L represent the length parallel to the wall and W the width perpendicular to it. The area of the pen is A = LW, and the fencing constraint is L + 2W = 100. This is a constrained optimization problem because the area must be maximized while the total fencing remains fixed.Lagrange multipliers provide a systematic way to incorporate the constraint into the optimization process. Geometrically, the maximum occurs where a level curve of the area function is tangent to the constraint line. At that point, both curves have the same tangent direction, so their gradients are parallel. This gives the condition\begin{equation*}\nabla A = \lambda \nabla g\end{equation*}where g(L, W) = L + 2W - 100 and &#x1D706; is the Lagrange multiplier. Next, taking partial derivatives of the area function and constraint with respect to L and W yields the equations\begin{equation*}W = \lambda, \qquad L = 2\lambda,\end{equation*}So, L = 2W, substituting it into the constraint gives\begin{equation*}W=25, \qquad L=50.\end{equation*}The rectangular pen with maximum area has a length of 50 meters and a width of 25 meters. This example shows how Lagrange multipliers connect algebraic optimization, geometric interpretation, and physical constraints in a single method.

Lagrange Multipliers: One Constraint

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optimizing a fence pen with lagrange multipliers

Lagrange multipliers help solve constrained optimization problems. In these problems, you maximize or minimize a quantity while keeping a fixed condition. A common example is a rectangular pen built against a barn wall using 100 meters of fencing.

The barn wall makes one side of the pen, so only the other three sides need fencing. The goal is to find the dimensions that give the greatest area. Let L be the length parallel to the wall and W be the width perpendicular to the wall. Then the area is A = LW, and the fencing rule is L + 2W = 100.

This setup is a constrained optimization problem because the area must be maximized while the fencing total stays fixed. Lagrange multipliers provide a systematic way to include that rule in the optimization process. Geometrically, the maximum happens where a level curve of the area function is tangent to the constraint line. At that point, the two curves have the same tangent direction, so their gradients are parallel.

That idea leads to the condition ∇A = λ∇g, where g(L, W) = L + 2W - 100 and λ is the Lagrange multiplier. Taking partial derivatives of the area function and the constraint with respect to L and W gives the equations W = λ and L = 2λ. This means L = 2W. Substituting into the constraint gives W = 25 and L = 50.

So, the rectangular pen with the maximum area has a length of 50 meters and a width of 25 meters. This example shows how Lagrange multipliers connect algebraic optimization, geometry, and a real-world fencing limit in one method.

Optimizing a Fence Pen with Lagrange Multipliers

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Lagrange multipliers help solve constrained optimization problems. In these problems, you maximize or minimize a quantity while keeping a fixed condition. A common example is a rectangular pen built against a barn wall using 100 meters of fencing. The barn wall makes one side of the pen, so only the other three sides need fencing. The goal is to find the dimensions that give the greatest area. Let L be the length parallel to the wall and W be the width perpendicular to the wall. Then the area...

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Learn how Lagrange multipliers solve a fenced pen problem by maximizing area under one constraint, using gradients, level curves, and a real example.

Optimizing a Fence Pen with Lagrange Multipliers | JoVE K12

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Optimizing a Fence Pen with Lagrange Multipliers

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