Question

find two positive numbers whose product is 323 and whose sum is a minimum.
the two numbers are
(type an exact answer, using radicals as needed. use a comma to separate answers as needed.)

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 \(1-\frac{323}{x^{2}}=0\).

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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:

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