Answer
(a) The Lagrange equations:
$2x = 4\lambda $, ${\ \ \ }$ $4y = - 6\lambda $.
(b) We show that we may assume that $x$ and $y$ are nonzero.
(c) Using the Lagrange equations we show that $y = - \frac{3}{4}x$.
(d) There is a unique critical point $P = \left( {\frac{{50}}{{17}}, - \frac{{75}}{{34}}} \right)$.
(e) $P$ corresponds to a minimum value of $f$.
Work Step by Step
(a) We have the function $f\left( {x,y} \right) = {x^2} + 2{y^2}$ and the constraint $g\left( {x,y} \right) = 4x - 6y - 25 = 0$.
Using Theorem 1, the Lagrange condition $\nabla f = \lambda \nabla g$ yields
$\left( {2x,4y} \right) = \lambda \left( {4, - 6} \right)$
Hence, $2x = 4\lambda $, ${\ }$ $4y = - 6\lambda $.
(b) In part (a) we obtain the Lagrange equations:
$2x = 4\lambda $, ${\ \ \ \ }$ $4y = - 6\lambda $
If $x=0$, then $\lambda = 0$. It follows that $y=0$.
Similarly, if $y=0$, then $\lambda = 0$. It follows that $x=0$.
Thus, if $x=0$ or $y=0$, then the Lagrange equations give $x=y=0$. However, $\left( {0,0} \right)$ does not satisfy the constraint $g\left( {x,y} \right) = 4x - 6y - 25 = 0$. Hence, we may assume that $x$ and $y$ are nonzero.
(c) In part (a) we obtain the Lagrange equations:
$2x = 4\lambda $, ${\ \ \ \ }$ $4y = - 6\lambda $
So, $\lambda = \frac{1}{2}x = - \frac{2}{3}y$. Hence, $y = - \frac{3}{4}x$.
(d) In part (c) we obtain $y = - \frac{3}{4}x$. Substitute it in the constraint equation gives
$g\left( {x, - \frac{3}{4}x} \right) = 4x + \frac{9}{2}x - 25 = 0$
$x = \frac{{50}}{{17}}$
Using $y = - \frac{3}{4}x$, we obtain $y = - \frac{{75}}{{34}}$. So, there is a unique critical point $P = \left( {\frac{{50}}{{17}}, - \frac{{75}}{{34}}} \right)$.
(e) Referring to Figure 13, as $\left( {x,y} \right)$ moves away from $P$ along the line $g\left( {x,y} \right) = 0$, either to the right or to the left, $f\left( {x,y} \right)$ increases. So, we conclude that $P$ corresponds to a minimum value of $f$.
