Answer to Question #202524 in Calculus for Sarita bartwal

Question #202524

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


1
Expert's answer
2021-06-03T15:28:11-0400

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



g(x,y)=x2+y2=1g(x, y)=x^2+y^2=1

Using Lagrange multipliers,



fx=λgx,fy=λgy,g(x,y)=1f_x=\lambda g_x, f_y=\lambda g_y, g(x, y)=1

which become



2x=2xλ2x=2x\lambda1=2yλ1=2y\lambdax2+y2=1x^2+y^2=1

If x=0,x=0, then y2=1=>y=±1.y^2=1=>y=\pm1.

If λ=1,\lambda=1, then y=12,y=\dfrac{1}{2}, and x=±32x=\pm\dfrac{\sqrt{3}}{2}

Therefore ff has possible extreme values at the points (0,1),(0,1),(32,12),(0, -1), (0, 1),( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}), and (32,12).(\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}). Evaluating ffat these four points, we find that 



f(0,1)=1f(0, -1)=-1f(0,1)=1f(0, 1)=1f(32,12)=54f( -\dfrac{\sqrt{3}}{2}, \dfrac{1}{2})=\dfrac{5}{4}f(32,12)=54f( \dfrac{\sqrt{3}}{2}, \dfrac{1}{2})=\dfrac{5}{4}

Therefore the maximum value of ff on the circle x2+y2=1x^2+y^2=1 is



f(±32,12)=54,f( \pm\dfrac{\sqrt{3}}{2}, \dfrac{1}{2})=\dfrac{5}{4},

the minimum value of ff on the circle x2+y2=1x^2+y^2=1 is



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

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