Answer to Question #341053 in Calculus for katie

Question #341053

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





subject to the constraint





y − x2 = 2

1
Expert's answer
2022-05-17T11:13:17-0400

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 "\u0393" describing the constraint is "\u0393=\\left \\{ \\left ( x,2+x^{2} \\right ): x\\in \\R \\right \\}" .On the set "\u0393" 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_{\u0393 }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_{\u0393}z(x,y)=13."


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