Find two positive numbers whose product is 121 and whose sum is a minimum.

Question

Quizplus · Accepted Answer

We can write the given number as $11eaae29\_f512\_bf7a\_87fe\_c7da465412f4\_TB2067\_11=11	imes10^{16}+eaae29\_f512\_bf7a\_87fe\_c7da465412f4\_TB2067\_11$. Let the two numbers be $x$ and $y$. We want to minimize $x+y$. By AM-GM inequality,
$$x+y\geq2\sqrt{xy}=2\sqrt{(11	imes10^{16}+eaae29\_f512\_bf7a\_87fe\_c7da465412f4\_TB2067\_11)}.$$ 
Now it is enough to minimize $11	imes10^{16}+eaae29\_f512\_bf7a\_87fe\_c7da465412f4\_TB2067\_11$ which is greater than $11	imes10^{16}$. Therefore, $x+y$ is minimized when $x=y=\sqrt{11	imes10^{16}+eaae29\_f512\_bf7a\_87fe\_c7da465412f4\_TB2067\_11}/2$. Hence the answer is D.

Find Two Positive Numbers Whose Product Is 121 and Whose