MHT CET · Maths · Application of Derivatives
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 & Solution
Correct Answer
(D) \((9,1)\)
Step-by-step Solution
Detailed explanation
Let \(x\) and \(y\) be the two parts of the number 10 .
\(\therefore\) \(x+y=10\ldots\) (i)
According to the question, Let
\(\begin{aligned} A &=2 x+y^{2} \\ &=2 x+(10-x)^{2} \\ &=2 x+100+x^{2}-20 x \\ &=x^{2}-18 x+100 \end{aligned}\)
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 \(y=1\)
\(\therefore (x, y)=(9,1)\)
\(\therefore\) \(x+y=10\ldots\) (i)
According to the question, Let
\(\begin{aligned} A &=2 x+y^{2} \\ &=2 x+(10-x)^{2} \\ &=2 x+100+x^{2}-20 x \\ &=x^{2}-18 x+100 \end{aligned}\)
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 \(y=1\)
\(\therefore (x, y)=(9,1)\)
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