2. find two positive numbers with product 200 such that the sum of one number and twice the second number is…

Explanation:

Step1: Let the two numbers be $x$ and $y$.

We know that $xy = 200$, so $y=\frac{200}{x}$. Let $S=x + 2y$. Substitute $y=\frac{200}{x}$ into $S$, we get $S=x+2\times\frac{200}{x}=x+\frac{400}{x}$, where $x>0$.

Step2: Find the derivative of $S$ with respect to $x$.

Using the power - rule, if $S(x)=x+\frac{400}{x}=x + 400x^{-1}$, then $S^\prime(x)=1-400x^{-2}=1-\frac{400}{x^{2}}$.

Step3: Set the derivative equal to zero to find critical points.

Set $S^\prime(x)=0$, so $1-\frac{400}{x^{2}} = 0$. Then $\frac{400}{x^{2}}=1$, and $x^{2}=400$. Since $x>0$, we have $x = 20$.

Step4: Find the second - derivative of $S$ to confirm it's a minimum.

$S^{\prime\prime}(x)=\frac{800}{x^{3}}$. When $x = 20$, $S^{\prime\prime}(20)=\frac{800}{20^{3}}=\frac{800}{8000}=\frac{1}{10}>0$. So $x = 20$ gives a minimum.

Step5: Find the value of $y$.

Since $y=\frac{200}{x}$ and $x = 20$, then $y = 10$.

Answer:

The two positive numbers are 20 and 10.