MHT CET · Maths · Application of Derivatives
10 is divided into two parts such that the sum of double of the first and square of the other is minimum, then the numbers are respectively
- A 9,1
- B 8,2
- C 6,4
- D 7,3
Answer & Solution
Correct Answer
(A) 9,1
Step-by-step Solution
Detailed explanation
Let the two parts of 10 be \(x\) and \((10-x)\).
\(
f(x)=2(10-x)+x^2=x^2-2 x+20
\)
\(f^{\prime}(x)=2 x-2\) and when \(f^{\prime}(x)=0\), we get \(x=1\)
\(
\mathrm{f}^{\prime \prime}(\mathrm{x})=2>0
\)
\(\therefore \mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=1\).
Thus the parts are 1,9 .
\(
f(x)=2(10-x)+x^2=x^2-2 x+20
\)
\(f^{\prime}(x)=2 x-2\) and when \(f^{\prime}(x)=0\), we get \(x=1\)
\(
\mathrm{f}^{\prime \prime}(\mathrm{x})=2>0
\)
\(\therefore \mathrm{f}(\mathrm{x})\) is minimum at \(\mathrm{x}=1\).
Thus the parts are 1,9 .
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