Answer
f(x) does not have an absolute minimum or maximum
Work Step by Step
$f(x,y)=8x^{2}y$
The constraint becomes $3x-y-9=0$ with $g(x,y)=3x-y-9$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=8x^{2}y-\lambda(3x-y-9)=x^{2}y-3\lambda x - \lambda y + 9\lambda$
$F_{x}(x,y,\lambda)=2xy-\lambda$
$F_{y}(x,y,\lambda)=x^{2}-\lambda$
$F_{\lambda}(x,y,\lambda)=-3x-y+9$
$(1) 2xy-\lambda =0 \rightarrow \lambda =2xy$
$(2) x^{2} -\lambda =0 \rightarrow \lambda =x^{2}$
$(3) -3x-y+9=0$
Set the expressions equal: $x^{2}=2xy$
$x=2y$
Substitute for x=2y in Equation (3)
$-6y-y+9=0$
$y=\frac{9}{7} \rightarrow x=\frac{18}{7}$
The value of is $f(\frac{18}{7},\frac{9}{7})\approx68.01$
Let $y=1, x=\frac{10}{3}$ then $f(\frac{10}{3},1)\approx88.89$. Because a nearby point has a value larger than 68.01, the value 68.01 is probably not a maximum.
Let's try with a different nearby point $y=1.5,x=3.5$ then $f(3.5,1.5)=147$
Since the two nearby points have a larger value compared to 68.01, f(x) does not have an absolute minimum or maximum.
