find two positive numbers whose product is 323 and whose sum is a minimum.\nthe two numbers are \n(type an…

Explanation:

Step1: Define variables

Let the two positive numbers be (x) and (y). Given (xy = 323), so (y=\frac{323}{x}). The sum (S=x + y=x+\frac{323}{x}), (x>0).

Step2: Find the derivative of the sum function

Differentiate (S(x)) with respect to (x). Using the power rule ((x^n)^\prime=nx^{n - 1}), (S^\prime(x)=1-\frac{323}{x^{2}}).

Step3: Find the critical points

Set (S^\prime(x) = 0), then (1-\frac{323}{x^{2}}=0). [ \begin{align*} \frac{x^{2}-323}{x^{2}}&=0\ x^{2}-323&=0\ x^{2}&=323\ x&=\sqrt{323}=\sqrt{17\times19}\text{ (since }x > 0) \end{align*} ]

Step4: Check the second - derivative

Differentiate (S^\prime(x)) to get (S^{\prime\prime}(x)=\frac{646}{x^{3}}). When (x = \sqrt{323}), (S^{\prime\prime}(\sqrt{323})=\frac{646}{(\sqrt{323})^{3}}>0), so (S(x)) has a minimum at (x=\sqrt{323}). When (x=\sqrt{323}), (y=\frac{323}{\sqrt{323}}=\sqrt{323}).

Answer:

(\sqrt{323},\sqrt{323})