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Question Number 204152 by universe last updated on 07/Feb/24
the maximum value of f(x,y) = xy−x^3 y^2 attained over the square 0≤x≤1;0≤y≤1 is
$$\:\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{xy}−\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} \\ $$$$\:\mathrm{attained}\:\mathrm{over}\:\mathrm{the}\:\mathrm{square}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1};\mathrm{0}\leqslant\mathrm{y}\leqslant\mathrm{1}\:\mathrm{is} \\ $$
Answered by ajfour last updated on 08/Feb/24
f(x,y)=z=t−t^3 (dz/dt)=1−3t^2 =0 ⇒ t=(1/( (√3))) z_(max) =(2/( (√(27)))) ≈ 0.38490
$${f}\left({x},{y}\right)={z}={t}−{t}^{\mathrm{3}} \\ $$$$\frac{{dz}}{{dt}}=\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}} =\mathrm{0}\:\:\Rightarrow\:\:{t}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${z}_{{max}} =\frac{\mathrm{2}}{\:\sqrt{\mathrm{27}}}\:\approx\:\mathrm{0}.\mathrm{38490} \\ $$$$ \\ $$
Answered by mr W last updated on 08/Feb/24
f(1,y)=y−y^2 =(1/4)−(y−(1/2))^2 f(1,y)_(max) =(1/4)=0.25 f(x,1)=x−x^3 ((df(x,1))/dx)=1−3x^2 =0 ⇒x=(1/( (√3))) f(x,1)_(max) =(1/( (√3)))−(1/(3(√3)))=((2(√3))/9)≈0.3849 ⇒f(x,y)_(max) =((2(√3))/9) ✓
$${f}\left(\mathrm{1},{y}\right)={y}−{y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}−\left({y}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${f}\left(\mathrm{1},{y}\right)_{{max}} =\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0}.\mathrm{25} \\ $$$${f}\left({x},\mathrm{1}\right)={x}−{x}^{\mathrm{3}} \\ $$$$\frac{{df}\left({x},\mathrm{1}\right)}{{dx}}=\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${f}\left({x},\mathrm{1}\right)_{{max}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\approx\mathrm{0}.\mathrm{3849} \\ $$$$\Rightarrow{f}\left({x},{y}\right)_{{max}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\:\checkmark \\ $$
Commented by ajfour last updated on 08/Feb/24
think i misread the question phone screen was not clean but this gave you some clue!
$${think}\:{i}\:{misread}\:{the}\:{question}\:{phone}\:{screen} \\ $$$${was}\:{not}\:{clean}\:{but}\:{this}\:{gave}\:{you}\:{some} \\ $$$${clue}! \\ $$
Commented by mr W last updated on 08/Feb/24
how i came to the path: since there is no solution for (∂f/∂x)=0 ∧ (∂f/∂y)=0 except at x=y=0, that means there is no extremum and we should search the maximum on the boundary: x=0 or x=1 or y=0 or y=1. x=0 or y=0 gives f=0 x=1 gives f_(max) =(1/4) y=1 gives f_(max) =((2(√3))/9) ⇒f_(max) =((2(√3))/9) for 0≤x, y≤1
$${how}\:{i}\:{came}\:{to}\:{the}\:{path}: \\ $$$${since}\:{there}\:{is}\:{no}\:{solution}\:{for}\: \\ $$$$\frac{\partial{f}}{\partial{x}}=\mathrm{0}\:\wedge\:\frac{\partial{f}}{\partial{y}}=\mathrm{0}\:{except}\:{at}\:{x}={y}=\mathrm{0}, \\ $$$${that}\:{means}\:{there}\:{is}\:{no}\:{extremum} \\ $$$${and}\:{we}\:{should}\:{search}\:{the} \\ $$$${maximum}\:{on}\:{the}\:{boundary}: \\ $$$${x}=\mathrm{0}\:{or}\:{x}=\mathrm{1}\:{or}\:{y}=\mathrm{0}\:{or}\:{y}=\mathrm{1}. \\ $$$${x}=\mathrm{0}\:{or}\:{y}=\mathrm{0}\:{gives}\:{f}=\mathrm{0} \\ $$$${x}=\mathrm{1}\:{gives}\:{f}_{{max}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}=\mathrm{1}\:{gives}\:{f}_{{max}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$$$\Rightarrow{f}_{{max}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\:{for}\:\mathrm{0}\leqslant{x},\:{y}\leqslant\mathrm{1} \\ $$
Commented by universe last updated on 08/Feb/24
thanks sir
$${thanks}\:{sir} \\ $$

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