# Answer in Calculus for Sarita bartwal #204491

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

Ans:-

Solve equations “u2207f= u03bb u2207g 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=2xlambda”“1=2ylambda”“x^2+y^2=1”

If “x=0,” then “y^2=1=>y=pm1.”

If “lambda=1,” then “y=dfrac{1}{2},” and “x=pmdfrac{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( pmdfrac{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”