Two positive numbers x and y are such that (x | vedclass

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(A) Given that $x+y=60$,so $x=60-y$. Let $f(y) = x y^3 = (60-y) y^3 = 60 y^3 - y^4$. To find the maximum,we differentiate $f(y)$ with respect to $y$:

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$15, 45$

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(A) Given that $x+y=60$,so $x=60-y$. Let $f(y) = x y^3 = (60-y) y^3 = 60 y^3 - y^4$. To find the maximum,we differentiate $f(y)$ with respect to $y$: $f'(y) = \frac{d}{dy}(60 y^3 - y^4) = 180 y^2 - 4 y^3$. Setting $f'(y) = 0$ for critical points: $4 y^2(45 - y) = 0$. Since $y$ is a positive number,$y=45$. Now,we check the second derivative: $f''(y) = 360 y - 12 y^2$. At $y=45$,$f''(45) = 360(45) - 12(45)^2 = 45(360 - 540) = 45(-180) Since the second derivative is negative,the function is maximum at $y=45$. Then $x = 60 - 45 = 15$. Thus,the numbers are $x=15$ and $y=45$.

Two positive numbers $x$ and $y$ are such that $(x+y)=60$ and $x y^3$ is maximum. Then the numbers $x$ and $y$ are respectively:

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