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Question Number 203527 by Mastermind last updated on 21/Jan/24

$$\mathrm{Find}\mathrm{the}\mathrm{maximum}\mathrm{value}\mathrm{of}\mathrm{the}\mathrm{function}$$$$\mathrm{f}(\mathrm{x},\mathrm{y})={\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2\mathrm{subject}\mathrm{to}\mathrm{the}\mathrm{condition}\mathrm{that}$$$${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2={\mathrm{c}}^2,\mathrm{where}\mathrm{c}\mathrm{is}\mathrm{the}\mathrm{constant}.$$$$$$$$\mathrm{Thank}\mathrm{you}\mathrm{in}\mathrm{advance}$$

Answered by aleks041103 last updated on 21/Jan/24

$$x^2=X$$$$y^2=Y$$$$z^2=Z$$$$c^2=C$$$$maxofXYZifX+Y+Z=C,X,Y,Z,C>0$$$$geom.mean⩽arithm.mean$$$$\Rightarrow \sqrt[3]{XYZ}⩽\frac{X+Y+Z}{3}=\frac{C}{3}(equalityifX=Y=Z)$$$$\Rightarrow XYZ⩽\frac{C^3}{27}$$$$\Rightarrow maxf(x,y,z)=x^2{y}^2{z}^{2}if{x}^2+y^2+z^2=c^2$$$$ismax\left(f\right)=\frac{c^6}{27}for\mid x\mid =\mid y\mid =\mid z\mid =\frac{\mid c\mid }{\sqrt{3}}$$

Commented by Mastermind last updated on 21/Jan/24

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{but}\mathrm{could}\mathrm{you}\mathrm{please}$$$$\mathrm{help}\mathrm{me}\mathrm{with}\mathrm{full}\mathrm{details}$$$$$$$$\mathrm{Thank}\mathrm{you}\mathrm{once}\mathrm{again}$$

Commented by aleks041103 last updated on 21/Jan/24

$$basicallyweusetheinequality$$$$GM⩽AM$$$$whereGMisthegeometricmean$$$$andAMisthearithmeticmean$$

Answered by deleteduser1 last updated on 21/Jan/24

$$c^2=x^2+y^2+z^2⩾3\sqrt[3]{x^2{y}^2{z}^2}\Rightarrow x^2{y}^2{z}^2⩽\frac{c^6}{27}$$

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