Question

Divide 10 into two parts such that the sum of twice of one part and square of the other is a minimum.

• A. 6,4
• B. 7,3
• C. 8,2
• D. 9, 1

Answer 1

The correct option is D 9, 1

Let one part be $x$

Therefore another part will be $10-x$.

Hence application of the condition gives us

$f\left(x\right)=x^2+2(10-x)$

$f^{\prime }\left(x\right)=2x-2$
$=0$

$x=1$

Hence the parts are $1,9$