Question

Divide $10$ into two parts such that the sum of double of the first and the square of the second is minimum

• A. $(6,4)$
• B. $(7,3)$
• C. $(8,2)$
• D. $(9,1)$

Answer 1

Answer: (D)

$(9,1)$

Let $x$ and $y$ be the two parts of the number $10$
$\therefore x+y=10$ ...... (i)

According to the question,
Let $A=2x+y^2$
$=2x+(10-x{)}^2$
$=2x+100+x^2-20x$
$=x^2-18x+100$

On differentiating w.r.t. $x$, we get
$\frac{dA}{dx}=2x-18$

For max or min of A,

Put $\frac{dA}{dx}=0=2x-18$

$\Rightarrow x=9$

Now, $\frac{d^{2}A}{d{x}^2}=2>0$ (min)
On putting $x=9$ in Eq. (i), we get
$y=1$
$\therefore (x,y)=(9,1)$