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Question Number 204152 by universe last updated on 07/Feb/24

$$\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{f}(\mathrm{x},\mathrm{y})=\mathrm{xy}-{\mathrm{x}}^3{\mathrm{y}}^2$$$$\mathrm{attained}\mathrm{over}\mathrm{the}\mathrm{square}0⩽\mathrm{x}⩽1;0⩽\mathrm{y}⩽1\mathrm{is}$$

Answered by ajfour last updated on 08/Feb/24

$$f(x,y)=z=t-t^3$$$$\frac{dz}{dt}=1-3t^2=0\Rightarrow t=\frac{1}{\sqrt{3}}$$$$z_{max}=\frac{2}{\sqrt{27}}\approx 0.38490$$$$$$

Answered by mr W last updated on 08/Feb/24

$$f(1,y)=y-y^2=\frac{1}{4}-{(y-\frac{1}{2})}^2$$$$f{(1,y)}_{max}=\frac{1}{4}=0.25$$$$f(x,1)=x-x^3$$$$\frac{df(x,1)}{dx}=1-3x^2=0\Rightarrow x=\frac{1}{\sqrt{3}}$$$$f{(x,1)}_{max}=\frac{1}{\sqrt{3}}-\frac{1}{3\sqrt{3}}=\frac{2\sqrt{3}}{9}\approx 0.3849$$$$\Rightarrow f{(x,y)}_{max}=\frac{2\sqrt{3}}{9}✓$$

Commented by ajfour last updated on 08/Feb/24

$$thinkimisreadthequestionphonescreen$$$$wasnotcleanbutthisgaveyousome$$$$clue!$$

Commented by mr W last updated on 08/Feb/24

$$howicametothepath:$$$$sincethereisnosolutionfor$$$$\frac{\partial f}{\partial x}=0\wedge \frac{\partial f}{\partial y}=0exceptatx=y=0,$$$$thatmeansthereisnoextremum$$$$andweshouldsearchthe$$$$maximumontheboundary:$$$$x=0orx=1ory=0ory=1.$$$$x=0ory=0givesf=0$$$$x=1gives{f}_{max}=\frac{1}{4}$$$$y=1gives{f}_{max}=\frac{2\sqrt{3}}{9}$$$$\Rightarrow f_{max}=\frac{2\sqrt{3}}{9}for0⩽x,y⩽1$$

Commented by universe last updated on 08/Feb/24

$$thankssir$$

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