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Question Number 118701 by bemath last updated on 19/Oct/20

$$Whatismaximumvalue$$$$x^3{y}^3+3xywhenx+y=8?$$

Commented by PRITHWISH SEN 2 last updated on 19/Oct/20

$$4144$$

Commented by bemath last updated on 19/Oct/20

$$x=y=4$$$${\left(16\right)}^3+3\times 4\times 4=4144$$$$$$

Answered by mr W last updated on 19/Oct/20

$${\left(xy\right)}^3+3\left(xy\right)$$$$={\left(\sqrt{xy}\right)}^6+3{\left(\sqrt{xy}\right)}^2$$$$⩽{\left(\frac{x+y}{2}\right)}^6+3{\left(\frac{x+y}{2}\right)}^2=4^6+3\times 4^2=4144$$

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