• anonymous
Use Lagrange multipliers to find the maximum or minimum values of f(x,y) subject to the constraint. f(x,y)= x^2+y, x^2-y^2=1
  • Stacey Warren - Expert
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  • katieb
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  • anonymous
We have the function \[f(x,y)=x^2+y\] and the constraint \[g(x,y)=x^2-y^2=1\] which is a hyperbola with the y=x and y=-x lines as asymptotes. Let's introduce the auxiliary function \[\Lambda(x,y,\lambda)=f(x,y)-\lambda(g(x,y)-1).\] Finding the critical points of this function is equivalent to finding the critical points of f(x,y) subject to the constraint. Compute the partial derivatives: \[\frac{\partial\Lambda}{\partial x},\quad\frac{\partial\Lambda}{\partial y},\quad\frac{\partial\Lambda}{\partial \lambda}\] then solve the system of linear equations \[\begin{array}{rcl}\frac{\partial\Lambda}{\partial x}&=&0,\\\frac{\partial\Lambda}{\partial y}&=&0,\\\frac{\partial\Lambda}{\partial\lambda}&=&0.\end{array}\] You'll get the solution \[x=\pm\frac{\sqrt{5}}{2},\quad y=-\frac{1}{2},\quad\lambda=-1.\] (You have to use that x cannot be zero on the hyperbola.) If you compute the value of f at the points \[(\frac{\sqrt{5}}{2},-\frac{1}{2})\textrm{ and }(-\frac{\sqrt{5}}{2},-\frac{1}{2})\] then you'll get \[f(\frac{\sqrt{5}}{2},-\frac{1}{2})=\frac{3}{4}\textrm{ and }f(-\frac{\sqrt{5}}{2},-\frac{1}{2})=\frac{3}{4}.\] You could use the Hessian matrix of Lambda the find out whether these point are maximum or minimum. They turn out to be minimums. Solution 2: A much easier way to find these points is use that\[x^2=1+y^2\textrm{ (this is the constraint)}\] and substitute the right-hand side to f. Then \[f(x(y),y)=1+y^2+y\] is just a single variable function (namely a parabola) which has a minimum at point y=-1/2. From the constraint the values of x will be\[x=\pm\frac{\sqrt{5}}{2}.\]

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