Answer to Question #204491 in Calculus for Sarita bartwal

Question

Find the extreme values of f(x,y)=x+y^2 on the surface 2x^2+y^2=1.

Answer 1

Ans:-

Solve equations $∇f= λ ∇g \ \ and \ \ \ g(x,y)=1$ using Lagrange multipliers Constraint:

$$g(x, y)=x^2+y^2=1$$

Using Lagrange multipliers,

$$f_x=\lambda g_x, f_y=\lambda g_y, g(x, y)=1$$

which become

$$2x=2x\lambda$$ $$1=2y\lambda$$ $$x^2+y^2=1$$

If $x=0,$ then $y^2=1=>y=\pm1.$

If $\lambda=1,$ then $y=\dfrac{1}{2},$ and $x=\pm\dfrac{\sqrt{3}}{2}$

Therefore $f$ has possible extreme values at the points $(0, -1), (0, 1),( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}),$ and $(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}).$ Evaluating $f$at these four points, we find that 

$$f(0, -1)=-1$$ $$f(0, 1)=1$$ $$f( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2})=\dfrac{5}{4}$$ $$f( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2})=\dfrac{5}{4}$$

Therefore the maximum value of $f$ on the circle $x^2+y^2=1$ is

$$f( \pm\dfrac{\sqrt{3}}{2}, \dfrac{1}{2})=\dfrac{5}{4},$$

the minimum value of $f$ on the circle $x^2+y^2=1$ is

$$f( 0,-1)=-1$$