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Question Number 118701 by bemath last updated on 19/Oct/20
What is maximum value   x^3 y^3  +3xy when x+y = 8?
$${What}\:{is}\:{maximum}\:{value} \\ $$$$\:{x}^{\mathrm{3}} {y}^{\mathrm{3}} \:+\mathrm{3}{xy}\:{when}\:{x}+{y}\:=\:\mathrm{8}? \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/Oct/20
4144
$$\mathrm{4144} \\ $$
Commented by bemath last updated on 19/Oct/20
x=y=4   (16)^3 +3×4×4 = 4144
$${x}={y}=\mathrm{4}\: \\ $$$$\left(\mathrm{16}\right)^{\mathrm{3}} +\mathrm{3}×\mathrm{4}×\mathrm{4}\:=\:\mathrm{4144} \\ $$$$ \\ $$
Answered by mr W last updated on 19/Oct/20
(xy)^3 +3(xy)  =((√(xy)))^6 +3((√(xy)))^2   ≤(((x+y)/2))^6 +3(((x+y)/2))^2 =4^6 +3×4^2 =4144
$$\left({xy}\right)^{\mathrm{3}} +\mathrm{3}\left({xy}\right) \\ $$$$=\left(\sqrt{{xy}}\right)^{\mathrm{6}} +\mathrm{3}\left(\sqrt{{xy}}\right)^{\mathrm{2}} \\ $$$$\leqslant\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{6}} +\mathrm{3}\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{6}} +\mathrm{3}×\mathrm{4}^{\mathrm{2}} =\mathrm{4144} \\ $$

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