Answer
$f(x,y)=3x^{2}+4y^{2}-xy-2$ has a minimum value subject to the constraint 2x+y=21, it is at the point $(4,\frac{17}{2})$. The value of $(4,\frac{17}{2})$ is 301.
Work Step by Step
$f(x,y)=3x^{2}+4y^{2}-xy-2$
The constraint becomes $2x+y-21=0$ with $g(x,y)=2x+y-21$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=3x^{2}+4y^{2}-xy-2-\lambda(2x+y-21)=3x^{2}+4y^{2}-xy-2-2\lambda x -\lambda y + 21\lambda$
$F_{x}(x,y,\lambda)=6x-y-2\lambda$
$F_{y}(x,y,\lambda)=8y-x-\lambda$
$F_{\lambda}(x,y,\lambda)=-2x-y+21$
$(1) 6x-y-2\lambda =0 \rightarrow \lambda =\frac{6x-y}{2}$
$(2) 8y-x-\lambda =0 \rightarrow \lambda =8y-x$
$(3) -2x-y+21=0$
Set the expressions for equal
$\frac{6x-y}{2}=8y-x$
$6x-y=16y-2x$
$x=\frac{17}{8}y$
Substitute for $x=\frac{17}{8}y$ in Equation (3)
$-2(\frac{17}{8}y)-y+21=0$
$y=4 \rightarrow x=\frac{17}{2}$
The value of f is $f(4,\frac{17}{2})=301$
Let y=9 and x=6 then f(6,9)=376 which is greater than 301.
Because a nearby point has a value greater than 301, the value 301 is probably not a maximum.
