Find the two positive numbers $x$ & $y$ such that their sum is $60$ and $x y^{3}$ is maximum

15

45

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30

30

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20

40

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10

50

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Solution

The correct option is C $15$ & $45$
Let one number be $x$
Hence the other number will be $(60 - x)$ .
Let
$K = x^{3} . (60 - x)$
Differentiating $K$ with respect to $x$ , we get
$\frac{d K}{d x}$
$= 3 x^{2} (60 - x) - x^{3} = 0$
Or
$x^{2} [180 - 3 x - x] = 0$
Or
$x = 0$ and $x = \frac{180}{4} = 45$ .
Now its given that the numbers are positive.
Hence $x = 0$ is ruled out.
Thus we get $x = 45$ .
Hence
$y = 15$ .
Therefore the numbers are $45, 15$ .

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Class-interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	Total
Frequency	$5$	$x$	$20$	$15$	$y$	$5$	$60$

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