Question

find two numbers a and b, with a ≤ b, whose difference is 50 and whose product is minimized.
a =
b =

Explanation:

Step1: Express \( B \) in terms of \( A \)

Given \( B - A=50 \), so \( B = A + 50 \).

Step2: Define the product function

Let \( P(A)=A\times B=A(A + 50)=A^{2}+50A \).

Step3: Find the derivative of the product function

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), \(P^\prime(A)=2A+50\).

Step4: Set the derivative equal to zero to find critical points

Set \(P^\prime(A)=0\), then \(2A+50 = 0\). Solving for \(A\):

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Step5: Find \( B \)

Since \(B = A + 50\), when \(A=-25\), \(B=-25 + 50=25\).

Step6: Verify it's a minimum (using the second - derivative test)

Find the second - derivative \(P^{\prime\prime}(A)=(2A + 50)^\prime=2>0\). Since \(P^{\prime\prime}(A)>0\), the function \(P(A)\) has a minimum at \(A=-25\).

Answer:

\(A=-25\), \(B = 25\)