Answer to Question #303646 in Calculus for Nofiupelumi

Question

Find the minimum and maximum values of f(x,y,z)=8x^2-2y subject to x^2+y^2=1

Answer 1

We are asked for the extreme values of $f$ subject to the constraint $g(x,y,z)=x^2+y^2=1.$

Using Lagrange multipliers, we solve the equations $\nabla f(x,y)=\lambda\nabla g(x, y)$ and $g(x,y)=1,$

which can be written as

$$f_x=\lambda g_x, f_y=\lambda g_y, x^2+y^2=1$$

$$16x=\lambda(2x), -2=\lambda(2y), x^2+y^2=1$$

We have $x=0$ or $\lambda=8.$

If $x=0,$ then $y=-1$ or $y=1.$

If $\lambda=8,$ then $y=-1/8$ and $x=\pm\dfrac{3\sqrt{7}}{8}$

$$f(0, -1)=8(0)^2-2(-1)=2$$

$$f(0, 1)=8(0)^2-2(1)=-2$$

$$f(-\dfrac{3\sqrt{7}}{8}, -\dfrac{1}{8})=8(\dfrac{-3\sqrt{7}}{8})^2-2(-\dfrac{1}{8})=\dfrac{65}{8}$$

$$f(\dfrac{3\sqrt{7}}{8}, -\dfrac{1}{8})=8(\dfrac{3\sqrt{7}}{8})^2-2(-\dfrac{1}{8})=\dfrac{65}{8}$$

Therefore the maximum value of on the circle $x^2+y^2=1$ is $f(\pm\dfrac{3\sqrt{7}}{8}, -\dfrac{1}{8})=\dfrac{65}{8} ,$ and the

minimum value is $f(0, 1)=-2.$