## anonymous 5 years ago 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

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$x=\pm\frac{\sqrt{5}}{2}.$