Answer
Function $f(x,y)=x^{3}y^{4}$ subject to constraint $g(x,y)=3x+3y-42$ has a maximum value of 884736 when x=6, y=8.
Work Step by Step
We are given $f(x,y)=x^{3}y^{4}$
where a unit of x cost is $\$3$, a unit of y cost is $\$3$ and $\$42$ is available, so that the combination of x and y is represented by the point $(x,y)=3x+3y=42 \rightarrow g(x,y)=3x+3y-42$
$F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{3}y^{4}-\lambda(3x+3y-42)=x^{3}y^{4}-3\lambda x - 3\lambda y + 42\lambda$
$F_{x}(x,y,\lambda)=3x^{2}y^{4}-3\lambda$
$F_{y}(x,y,\lambda)=4x^{3}y^{3}-3\lambda$
$F_{\lambda}(x,y,\lambda)=-3x-3y+42$
$(1) 3x^{2}y^{4}-3\lambda=0 \rightarrow \lambda =x^{2}y^{4}$
$(2) 4x^{3}y^{3}-3\lambda =0 \rightarrow \lambda =\frac{4x^{3}y^{3}}{3}$
$(3) -3x-3y+42=0$
Set the expressions equal: $x^{2}y^{4}=\frac{4x^{3}y^{3}}{3}$
$4x^{3}y^{3}=3x^{2}y^{4}$
$x=\frac{3}{4}y$
Substitute for $x=\frac{3}{4}y$ in Equation (3)
$-3(\frac{3}{4}y)-3y+42=0$
$y=8 \rightarrow x=6$
The value of is $f(6,8)=884736$
Let x=5 and y=9 then $f(5;9)=820125$. Because a nearby point has a value smaller than 884736, the value 884736 is probably not a minimum.
Function $f(x,y)=x^{3}y^{4}$ subject to constraint $g(x,y)=3x+3y-42$ has a maximum value of 884736 when x=6, y=8.
