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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

$$whatdoyougetwiththisexample:$$$$x=-135$$$$y=100$$$$whichfulfills3x+4y=5,but$$$$x^2{y}^3=18225000000>>\frac{3}{16}$$$$$$$$withy=x^3+3$$$$\frac{dy}{dx}=3x^2=0\Rightarrow x=0\Rightarrow y=3$$$$\Rightarrow youcannotsay{y}_{max}or{y}_{min}=3$$

Commented by liberty last updated on 09/May/21

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

Commented by liberty last updated on 09/May/21

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

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

$$aty=\frac{3}{4}itisonlya‘‘{local}^{″}maximum.$$$$it{doesn}^{\prime }tmean{x}^2{y}^3⩽\frac{3}{16}.$$

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