## Calculus with Applications (10th Edition)

Function $f(x,y)=x^{2}y^{3}$ subject to constraint $g(x,y)=2x+y-80$ has a maximum value of 28311552 when x=16, y=48
We are given $f(x,y)=x^{2}y^{3}$ where a unit of x cost is $\$2$, a unit of cost y is$\$1$ and $\$80$is available, so that the combination of x and y are represented by the point$(x,y)=2x+y=80 \rightarrow g(x,y)=2x+y-80F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{2}y^{3}-\lambda(2x+y-80)=x^{2}y^{3}-2\lambda x - \lambda y + 80\lambdaF_{x}(x,y,\lambda)=2xy^{3}-2\lambdaF_{y}(x,y,\lambda)=3x^{2}y^{2}-\lambdaF_{\lambda}(x,y,\lambda)=-2x-y+80(1) 2xy^{3}-2\lambda=0 \rightarrow \lambda =xy^{3}(2) 3x^{2}y^{2}-\lambda =0 \rightarrow \lambda =3x^{2}y^{2}(3) -2x-y+80=0$Set the expressions equal:$xy^{3}=3x^{2}y^{2}y=3x$Substitute for x=y in Equation (3)$-2x-3x+80=0x=16 \rightarrow y=48$The value of is$f(16,48)=28311552$Let x=15 and y=30 then$f(15,30)=6075000$. Because a nearby point has a value smaller than 28311552, the value 28311552 is probably not a minimum. Function$f(x,y)=x^{2}y^{3}$subject to constraint$g(x,y)=2x+y-80\$ has a maximum value of 28311552 when x=16, y=48.