Answer to Question #341053 in Calculus for katie

Question

Find the maximum value of z=6x−3x2 +2y

subject to the constraint

y − x2 = 2

Answer 1

ANSWER The maximum value of $z=6x-3x^{2}+2y$ subject to the constraint $y-x^{2}=2$ is $z(3,11)=13$ .

EXPLANATION

The set $Γ$ describing the constraint is $Γ=\left \{ \left ( x,2+x^{2} \right ): x\in \R \right \}$ .On the set $Γ$ the function $z(x,y)=6x-3x^{2}+2y$ has the form $z(x,2+x^{2})=6x-3x^{2}+2(2+x^{2})=-x^{2}+6x+4$ for all $x\in\R$ . Denote $h(x)=z(x,2+x^{2}).$ Therefore ,

$max_{Γ }z(x,y) =max_{\R}h(x)$ .

Since $h'(x)=-2x+6$ , then $h'(x)>0$ if $x\in (-\infty,3)$ and $h'(x)<0$ if $x\in(3,+\infty)$. $h'(3)=0$ . Since $h$ increases on the interval $(-\infty,3)$ and decreases on the interval $(3,+\infty)$ then $max_{(-\infty,3)}h(x)=h(3)=max_{(3,+\infty)}h(x)$ . So $max_{\R}h(x)=h(3)=-9+18+4=13$ $\Rightarrow max_{Γ}z(x,y)=13.$