## Calculus: Early Transcendentals 8th Edition

Maximum:$f(3,1)=10$, Minimum: $f(-3,-1)=-10$
Use Lagrange Multipliers Method: $\nabla f(x,y)=\lambda \nabla g(x,y)$ This yields $\nabla f=\lt 3,1 \gt$ and $\lambda \nabla g=\lambda \lt 2x,2y \gt$ Using the constraint condition $x^2+y^2=10$ we get, $3=\lambda 2x, 1=\lambda 2y$ After solving, we get $x=\pm 3$ Since, $g(x,y)=x^2+y^2=10$ gives $y=\pm 1$ Hence, Maximum:$f(3,1)=10$, Minimum: $f(-3,-1)=-10$