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The sum of two numbers is 20 . If the product of the square of one number and cube of the other is maximum, then the numbers are :
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The correct answer is:
12,8
Let the numbers be $x, y$.
$\therefore \quad x+y=20 \Rightarrow y=20-x$
Let $S=x^3(20-x)^2$
On differentiating w.r.t $x$, we get
$\frac{d S}{d x}=2 x^3(20-x)(-1)+3 x^2(20-x)^2$
For maximum or minimum, put $\frac{d S}{d x}=0$
$\Rightarrow \quad x^2(20-x)(-2 x+60-3 x)=0$
$\Rightarrow \quad x=0, x=12$ or $x=20$
$x$ cannot be 0 or 20
When $x=12$, then $y=20-12=8$
The required numbers are 12,8 .
$\therefore \quad x+y=20 \Rightarrow y=20-x$
Let $S=x^3(20-x)^2$
On differentiating w.r.t $x$, we get
$\frac{d S}{d x}=2 x^3(20-x)(-1)+3 x^2(20-x)^2$
For maximum or minimum, put $\frac{d S}{d x}=0$
$\Rightarrow \quad x^2(20-x)(-2 x+60-3 x)=0$
$\Rightarrow \quad x=0, x=12$ or $x=20$
$x$ cannot be 0 or 20
When $x=12$, then $y=20-12=8$
The required numbers are 12,8 .
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