Calculus with Applications (10th Edition)

$f(x,y)=12xy-x^{2}-3y^{2}$ has a maximum value subject to the constraint x+y=16; it is at the point (9,7). The value of (9,7) is 528.
$f(x,y)=12xy-x^{2}-3y^{2}$ The constraint becomes $x+y-16=0$ with $g(x,y)=x+y-16$ $F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=12xy-x^{2}-3y^{2}-\lambda(x+y-16)=12xy-x^{2}-3y^{2}-\lambda x - \lambda y + 16\lambda$ $F_{x}(x,y,\lambda)=12y-2x-\lambda$ $F_{y}(x,y,\lambda)=12x-6y-\lambda$ $F_{\lambda}(x,y,\lambda)=-x-y+16$ $(1) 12y-2x-\lambda=0 \rightarrow \lambda =12y-2x$ $(2) 12x-6y-\lambda =0 \rightarrow \lambda =12x-6y$ $(3) -x-y+16=0$ Set the expressions equal: $12y-2x=12x-6y$ $14x=18y$ $x=\frac{9}{7}y$ Substitute for x=y in Equation (3) $-\frac{9}{7}y-y+16=0$ $y=7 \rightarrow x=9$ The value of f is $f(9,7)=528$ Let x=8.9 and y=7.1 then $f(8.9;7.1)=527.84$. Because a nearby point has a value smaller than 528, the value 528 is probably not a minimum. $f(x,y)=12xy-x^{2}-3y^{2}$ has a maximum value subject to the constraint x+y=16; it is at the point (9,7). The value of (9,7) is 528.