## Calculus with Applications (10th Edition)

Function $f(x,y)=x^{4}y^{2}$ subject to the constraint $g(x,y)=2x+4y-60$ has a maximum value of 4000000 when x=20, y=5.
We are given $f(x,y)=x^{4}y^{2}$ where a unit of x cost is $\$2$, a unit of cost y is$\$4$ and $\$60$is available, so that the combination of x and y is represented by the point$(x,y)=2x+4y=60 \rightarrow g(x,y)=2x+4y-60F(x,y,\lambda)=f(x,y)-\lambda.g(x,y)=x^{4}y^{2}-\lambda(2x+4y-60)=x^{4}y^{2}-2\lambda x - 4\lambda y + 60\lambdaF_{x}(x,y,\lambda)=4x^{3}y^{2}-2\lambdaF_{y}(x,y,\lambda)=2x^{4}y-4\lambdaF_{\lambda}(x,y,\lambda)=-2x-4y+60(1) 4x^{3}y^{2}-2\lambda=0 \rightarrow \lambda =2x^{3}y^{2}(2) 2x^{4}y-4\lambda =0 \rightarrow \lambda =\frac{x^{4}y}{2}(3) -2x-4y+60=0$Set the expressions equal:$2x^{3}y^{2}=\frac{x^{4}y}{2}4x^{3}y^{2}=x^{4}yx=4y$Substitute for x=4y in Equation (3)$-2(4y)-4y+60=0y=5 \rightarrow x=20$The value of is$f(20,5)=4000000$Let x=19 and y=5.5 then$f(19,5.5)=3942210$. Because a nearby point has a value smaller than 4000000, the value 4000000 is probably not a minimum. Function$f(x,y)=x^{4}y^{2}$subject to constraint$g(x,y)=2x+4y-60\$ has a maximum value of 4000000 when x=20, y=5.