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Chapter 10: Problem 15

Use a Lagrange multiplier to find the curve \(x(y)\) of length \(L=\pi\) on the interval \([0,1]\) which maximizes the integral \(I=\int_{0}^{1} y(x) d x\) and pass through the points \((0,0)\) and \((1,0)\).

Short Answer

Expert verified
We formulate a function combining both the integral to be maximized and the constraint, then apply the Euler-Lagrange equation to extremize the function. Solving the differential equation and applying boundary conditions will yield the optimal curve.

Step by step solution

01

Problem Formulation

Firstly, let's write down the functional to be optimized which is \(I=\int_{0}^{1} y(x) dx\). The constraint is a fixed curve length, \(L=\pi\), which can be expressed as \(L=\int_{0}^{1} \sqrt{1+y'(x)^{2}} dx = \pi\). Using Lagrange multipliers, we combine those two equations into a function \(F\), which is \(F= y+\lambda(\sqrt{1+y'^2}-1)\), where \(\lambda\) is a yet undetermined Lagrange multiplier.
02

Deriving the Euler-Lagrange Equation

We are going to use the Euler-Lagrange equation to extremize the above functional. For a functional \(F\), the equation is \(\frac{\partial F}{\partial y} - \frac{d }{dx}(\frac{\partial F}{\partial y'}) = 0\). Apply this to \(F\) gives the equation \(1 -\frac{d }{dx} \frac{\lambda y'}{\sqrt{1+y'^2}} = 0\). After doing the derivatives and simplifying, the resulting differential equation is \(\frac{d }{dx} \frac{\lambda y'}{\sqrt{1+y'^2}} = 1\).
03

Solving the Differential Equation

Integrating each side with respect to \(x\), we obtain \(\lambda \sqrt{1+y'^2} = x + A\) where \(A\) is an integration constant. This is a first order nonlinear ordinary differential equation for the derivative \(y'(x)\). Solving this for \(y'(x)\) with the boundary condition \(y(0)=0\) will yield the final solution.
04

Determining the Lagrange Multiplier and Integrating Constant

The Lagrange multiplier \(\lambda\) and the integration constant \(A\) can be determined by using the second boundary condition \(y(1)=0\) and the constraint \(L=\pi\).
05

Final Step – Evaluating The Integral

Substituting the results of \(\lambda\) and \(A\) into the differential equation and then integrating it will yield the curve \(x(y)\) that maximizes the integral \(I\) under the given condition and through solving this, a conclusion to the exercise is found.

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