## Related questions with answers

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

$f ( x , y ) = x ^ { 2 } + y ^ { 2 } , \quad x ^ { 4 } + y ^ { 4 } = 1$

Solution

Verified$\textbf { Lagrange equations:}$

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

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

The Lagrange equations $\nabla f_p= \lambda \nabla g_p$ are

$\langle 2x,2y \rangle= \lambda \langle 4x^3,4y^3 \rangle \implies 2x= 4x^3 \lambda , \hspace{0.5cm} 2y= 4y^3 \lambda$

.

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

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

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