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Question Number 140550 by liberty last updated on 09/May/21

Commented by mr W last updated on 09/May/21

what do you get with this example: x=−135 y=100 which fulfills 3x+4y=5, but x^2 y^3 =18 225 000 000 >> (3/(16)) with y=x^3 +3 (dy/dx)=3x^2 =0 ⇒x=0 ⇒y=3 ⇒you can not say y_(max) or y_(min) =3

$${what}\:{do}\:{you}\:{get}\:{with}\:{this}\:{example}: \\ $$$${x}=−\mathrm{135} \\ $$$${y}=\mathrm{100} \\ $$$${which}\:{fulfills}\:\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{5},\:{but} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{3}} =\mathrm{18}\:\mathrm{225}\:\mathrm{000}\:\mathrm{000}\:>>\:\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$ \\ $$$${with}\:{y}={x}^{\mathrm{3}} +\mathrm{3} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{x}=\mathrm{0}\:\Rightarrow{y}=\mathrm{3} \\ $$$$\Rightarrow{you}\:{can}\:{not}\:{say}\:{y}_{{max}} \:{or}\:{y}_{{min}} =\mathrm{3} \\ $$

Commented by liberty last updated on 09/May/21

if your function y=x^3 +3 (dy/dx) =3x^2 = 0⇒x=0 then (d^2 y/dx^2 ) = 6x < 0 for max since 6x=0 for x = 0 so y=x^3 +3 have no max value

$$\mathrm{if}\:\mathrm{your}\:\mathrm{function}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} +\mathrm{3} \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\mathrm{3x}^{\mathrm{2}} \:=\:\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{0} \\ $$$$\mathrm{then}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=\:\mathrm{6x}\:<\:\mathrm{0}\:\mathrm{for}\:\mathrm{max} \\ $$$$\mathrm{since}\:\mathrm{6x}=\mathrm{0}\:\mathrm{for}\:\mathrm{x}\:=\:\mathrm{0}\:\mathrm{so}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} +\mathrm{3} \\ $$$$\mathrm{have}\:\mathrm{no}\:\mathrm{max}\:\mathrm{value} \\ $$

Commented by liberty last updated on 09/May/21

3x+4y=5⇒x=((5−4y)/3) f(y)=(((5−4y)/3))^2 y^3 f ′(y)=−(8/3)y^3 (((5−4y)/3))+3y^2 (((5−4y)/3))^2 =0 y^2 (((5−4y)/3))[−(8/3)y+3(((5−4y)/3))]=0 y^2 (((5−4y)/3))[((−20y+15)/3)]=0 y=0; (5/4) ; (3/4) ⇒f(y) decreasing for (3/4)<y<(5/4) and increasing for y<0 ∪ 0<y<(3/4) ∪ y>(5/4) so y=(3/4) is critical point

$$\mathrm{3x}+\mathrm{4y}=\mathrm{5}\Rightarrow\mathrm{x}=\frac{\mathrm{5}−\mathrm{4y}}{\mathrm{3}} \\ $$$$\mathrm{f}\left(\mathrm{y}\right)=\left(\frac{\mathrm{5}−\mathrm{4y}}{\mathrm{3}}\right)^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} \\ $$$$\mathrm{f}\:'\left(\mathrm{y}\right)=−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{y}^{\mathrm{3}} \left(\frac{\mathrm{5}−\mathrm{4y}}{\mathrm{3}}\right)+\mathrm{3y}^{\mathrm{2}} \left(\frac{\mathrm{5}−\mathrm{4y}}{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{y}^{\mathrm{2}} \left(\frac{\mathrm{5}−\mathrm{4y}}{\mathrm{3}}\right)\left[−\frac{\mathrm{8}}{\mathrm{3}}\mathrm{y}+\mathrm{3}\left(\frac{\mathrm{5}−\mathrm{4y}}{\mathrm{3}}\right)\right]=\mathrm{0} \\ $$$$\mathrm{y}^{\mathrm{2}} \left(\frac{\mathrm{5}−\mathrm{4y}}{\mathrm{3}}\right)\left[\frac{−\mathrm{20y}+\mathrm{15}}{\mathrm{3}}\right]=\mathrm{0} \\ $$$$\mathrm{y}=\mathrm{0};\:\frac{\mathrm{5}}{\mathrm{4}}\:;\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{y}\right)\:\mathrm{decreasing}\:\mathrm{for}\:\frac{\mathrm{3}}{\mathrm{4}}<\mathrm{y}<\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{and}\:\mathrm{increasing}\:\mathrm{for}\:\mathrm{y}<\mathrm{0}\:\cup\:\mathrm{0}<\mathrm{y}<\frac{\mathrm{3}}{\mathrm{4}}\:\cup\:\mathrm{y}>\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{so}\:\mathrm{y}=\frac{\mathrm{3}}{\mathrm{4}}\:\mathrm{is}\:\mathrm{critical}\:\mathrm{point} \\ $$

Commented by mr W last updated on 09/May/21

at y=(3/4) it is only a “local” maximum. it doesn′t mean x^2 y^3 ≤(3/(16)).

$${at}\:{y}=\frac{\mathrm{3}}{\mathrm{4}}\:{it}\:{is}\:{only}\:{a}\:``{local}''\:{maximum}. \\ $$$${it}\:{doesn}'{t}\:{mean}\:{x}^{\mathrm{2}} {y}^{\mathrm{3}} \leqslant\frac{\mathrm{3}}{\mathrm{16}}. \\ $$

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