Q64 / 150 · MHT CET 2012 · Mathematics
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$(8,2)$
- Correct answerD$(9,1)$
Solution
Let $x$ and $y$ be the two parts of the number 10 .
$\therefore$
$x+y=10\ldots$ (i)
According to the question, Let $\quad \begin{aligned} A &=2 x+y^{2} \\ &=2 x+(10-x)^{2} \\ &=2 x+100+x^{2}-20 x \\ &=x^{2}-18 x+100 \end{aligned}$
On differentiating w.r.t. $x$, we get $\frac{d A}{d x}=2 x-18$
For max or min of $A$, Put $\quad \frac{d A}{d x}=0=2 x-18$
$\Rightarrow$
$x=9$
Now, $\quad \frac{d^{2} A}{d x^{2}}=2>0 \quad(\min )$
On putting $x=9$ in Eq. (i), we get $y=1$
$\therefore \quad(x, y)=(9,1)$
Hence, the correct answer is option: 'D'.