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Question

Find the minimum and maximum values of the function subject to the given constraint.

f(x,y)=x2+y2,x4+y4=1f ( x , y ) = x ^ { 2 } + y ^ { 2 } , \quad x ^ { 4 } + y ^ { 4 } = 1

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 Lagrange equations:\textbf { Lagrange equations:}

Here f(x,y)=x2+y2f(x,y)=x^2+y^2 and the constraint curve is g(x,y)=0g(x, y) = 0, where g(x,y)=x4+y41g(x, y) = x^4+y^4-1 and

f=2x,2y,g=4x3,4y3.\nabla f= \langle 2x,2y \rangle, \hspace{0.5cm} \nabla g= \langle 4x^3,4y^3 \rangle.

The Lagrange equations fp=λgp\nabla f_p= \lambda \nabla g_p are

2x,2y=λ4x3,4y3    2x=4x3λ,2y=4y3λ\langle 2x,2y \rangle= \lambda \langle 4x^3,4y^3 \rangle \implies 2x= 4x^3 \lambda , \hspace{0.5cm} 2y= 4y^3 \lambda

.

 Solving for λ in terms of x and y :\textbf{ Solving for $\lambda$ in terms of $x$ and $y$ :}

Now λ0\lambda \neq 0, otherwise x=y=0x=y=0, which violate the constraint equation. Now 2x=4x3λ    x(12x2λ)=02x= 4x^3 \lambda \implies x(1-2x^2 \lambda)=0. Suppose x=0x=0, then from constraint equation we get y=±1y=\pm 1. Similarly 2y=4y3λ    y(12y2λ)=02y= 4y^3 \lambda \implies y(1-2y^2 \lambda)=0 and suppose y=0y=0 then x=±1x= \pm 1. Now consider the case where x,y0x,y \neq 0. Then we can write

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