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Two positive numbers have a sum of $15$. The sum of their squares is minimum when the numbers are
A$5$ and $10$
B$\tfrac{15}{2}$ and $\tfrac{15}{2}$
C$\tfrac{15}{4}$ and $\tfrac{45}{4}$
D$\tfrac{5}{2}$ and $\tfrac{25}{2}$
Answer & Solution
Correct answer: B. $\tfrac{15}{2}$ and $\tfrac{15}{2}$
1. Let the numbers be $x$ and $15-x$.
2. $S(x)=x^2+(15-x)^2=2x^2-30x+225$.
3. $S'(x)=4x-30=0 \Rightarrow x=\tfrac{15}{2}$.
4. $S''(x)=4>0$, so this is a minimum.
5. The numbers are $\tfrac{15}{2}$ and $15-\tfrac{15}{2}=\tfrac{15}{2}$.
_Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.20_
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