Answer
(a) The Lagrange equation gives $\lambda x = 1$ and $\lambda y = 2$.
(b) We show that $\lambda \ne 0$ and $y=2x$.
(c) The critical points are $\left( {1,2} \right)$ and $\left( { - 1, - 2} \right)$.
(d) The maximum value is $f\left( {1,2} \right) = 10$ and the minimum value is $f\left( { - 1, - 2} \right) = - 10$.
Work Step by Step
(a) We have the function $f\left( {x,y} \right) = 2x + 4y$ and the constraint $g\left( {x,y} \right) = {x^2} + {y^2} - 5 = 0$.
By Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ gives
$\left( {2,4} \right) = \lambda \left( {2x,2y} \right)$
Hence, $\lambda x = 1$ and $\lambda y = 2$.
(b) In part (a) we obtain $\lambda x = 1$ and $\lambda y = 2$. Since the right-hand sides are nonzero, this implies that $\lambda \ne 0$.
So, $\lambda = \frac{1}{x} = \frac{2}{y}$. It follows that $y=2x$.
(c) In part (b) we obtain $y=2x$. Substituting it in the constraint equation $g\left( {x,y} \right) = {x^2} + {y^2} - 5 = 0$ gives
${x^2} + 4{x^2} - 5 = 0$
So, the solution is $x = \pm 1$.
The critical points are $\left( {1,2} \right)$ and $\left( { - 1, - 2} \right)$.
(d) Evaluating $f$ at the critical points $\left( {1,2} \right)$ and $\left( { - 1, - 2} \right)$, we get $f\left( {1,2} \right) = 10$ and $f\left( { - 1, - 2} \right) = - 10$.
Thus, the maximum value is $f\left( {1,2} \right) = 10$ and the minimum value is $f\left( { - 1, - 2} \right) = - 10$.
