Question

find two positive numbers that satisfy the given requirements. (enter your answers as a comma - separated list.) the product is 192 and the sum of the first plus three times the second is a minimum.

Explanation:

Step1: Define variables

Let the two positive numbers be \(x\) and \(y\). Given \(xy = 192\), so \(x=\frac{192}{y}\). The function to minimize is \(S=x + 3y\). Substitute \(x\) into \(S\): \(S=\frac{192}{y}+3y\).

Step2: Find the derivative

Differentiate \(S(y)\) with respect to \(y\). Using the power rule \((u^{-n})^\prime=-nu^{-n - 1}\) and \((ay)^\prime=a\) (\(a\) is a constant), \(S^\prime(y)=-\frac{192}{y^{2}}+3\).

Step3: Set the derivative equal to zero

Set \(S^\prime(y) = 0\), then \(-\frac{192}{y^{2}}+3 = 0\).

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Step4: Find \(x\)

Since \(x=\frac{192}{y}\) and \(y = 8\), then \(x=\frac{192}{8}=24\).

Step5: Second - derivative test (optional for verification)

Differentiate \(S^\prime(y)\) to get \(S^{\prime\prime}(y)=\frac{384}{y^{3}}\). When \(y = 8\), \(S^{\prime\prime}(8)=\frac{384}{8^{3}}=\frac{384}{512}=\frac{3}{4}>0\). So \(S(y)\) has a minimum at \(y = 8\).

Answer:

\(24,8\)