If x+y=60,x0,y0,then the maximum value of x | vedclass

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(D) Given,$x+y=60$.Using the $AM \geq GM$ inequality for the terms $x, \frac{y}{3}, \frac{y}{3}, \frac{y}{3}$:$\frac{x + \frac{y}{3} + \frac{y}{3} + \

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$\frac{(45)^4}{3}$

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(D) Given,$x+y=60$.Using the $AM \geq GM$ inequality for the terms $x, \frac{y}{3}, \frac{y}{3}, \frac{y}{3}$:$\frac{x + \frac{y}{3} + \frac{y}{3} + \frac{y}{3}}{4} \geq \sqrt[4]{x \cdot \frac{y}{3} \cdot \frac{y}{3} \cdot \frac{y}{3}}$$\frac{x+y}{4} \geq \sqrt[4]{\frac{x y^3}{27}}$Substituting $x+y=60$:$\frac{60}{4} \geq \sqrt[4]{\frac{x y^3}{27}}$$15 \geq \sqrt[4]{\frac{x y^3}{27}}$Raising both sides to the power of $4$:$(15)^4 \geq \frac{x y^3}{27}$$x y^3 \leq 27 	imes (15)^4$$x y^3 \leq 3^3 	imes (3 	imes 5)^4$$x y^3 \leq 3^3 	imes 3^4 	imes 5^4$$x y^3 \leq 3^7 	imes 5^4$Since $\frac{(45)^4}{3} = \frac{(3^2 	imes 5)^4}{3} = \frac{3^8 	imes 5^4}{3} = 3^7 	imes 5^4$,the maximum value is $\frac{(45)^4}{3}$.

If $x+y=60$,$x>0$,$y>0$,then the maximum value of $x y^3$ is

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