Junior 2006/3
Prove that the product of two real numbers is maximum when the numbers are equal to each other while $a+b=$constant.
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- asif e elahi
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Re: Junior 2006/3
Use AM-GM inequality.
Re: Junior 2006/3
sorry,I have got a mistake.There is (a+b) after while.This is: while a+b=constant
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Re: Junior 2006/3
Please give the solution
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- asif e elahi
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Re: Junior 2006/3
Let $a+b=m$.Then $ab\leq (\frac{a+b}{2})^{2}=\frac{m^{2}}{4}$. Equality holds when $a=b$.
- Fatin Farhan
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Re: Junior 2006/3
Can also be done without using anything.
$$b=m-a$$
$$ab=a(m-a)=am-a^2$$
$$=\frac{m^2}{4} -a^2+ 2\frac{m}{2}a-\frac{m^2}{4}$$
$$=\frac{m^2}{4}- (a-\frac{m}{2})^2$$.
So, ab will be maximum if
$$a-\frac{m}{2}=0$$
$$a=\frac{m}{2}$$.
$$b=m-a=\frac{m}{2}$$
$$b=m-a$$
$$ab=a(m-a)=am-a^2$$
$$=\frac{m^2}{4} -a^2+ 2\frac{m}{2}a-\frac{m^2}{4}$$
$$=\frac{m^2}{4}- (a-\frac{m}{2})^2$$.
So, ab will be maximum if
$$a-\frac{m}{2}=0$$
$$a=\frac{m}{2}$$.
$$b=m-a=\frac{m}{2}$$
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