Problem 13 - Calculus Early Transcendentals

Question

Find positive numbers \(x\) and \(y\) satisfying the equation \(x y=12\) such that the sum \(2 x+y\) is as small as possible.

Step-by-Step Solution

Answer

Answer: x = √6 and y = 12/√6

Step 1: Express y in terms of x using the given equation

Let \(x\) and \(y\) be the positive numbers satisfying \(x y = 12\). We can express \(y\) in terms of \(x\) using this equation: \[ y = \frac{12}{x} \]

Step 2: Express the sum \(2x+y\) as a function of x

The sum that we want to minimize is \(2x+y\). Substitute the expression we found for \(y\) in terms of \(x\) into this sum: \[f(x) = 2x+\frac{12}{x}\]

Step 3: Find the critical points of f(x) by taking the derivative and setting it to 0

To minimize the function \(f(x)\), we need to find its critical points. To do this, take the derivative of \(f(x)\) with respect to \(x\) and set it equal to \(0\). Derivative of \(f(x)\): \[f'(x) = 2 - \frac{12}{x^2}\] Now, set \(f'(x)\) equal to 0 and solve for \(x\): \[2 - \frac{12}{x^2} = 0\]

Step 4: Solve the equation from Step 3 for x

Add \(\frac{12}{x^2}\) to both sides of the equation and then multiply both sides by \(x^2\): \[2x^2 = 12\] Divide both sides by 2: \[x^2 = 6\] Take the square root of both sides: \[x = \sqrt{6}\]

Step 5: Check if the critical point yields a local minimum

To determine whether the \(x\)-value we found results in a minimum value of the function \(f(x)\), we need to check the signs of the second derivative, \(f''(x)\): \[f''(x) = \frac{24}{x^3}\] Since \(x = \sqrt{6}\) and the second derivative \(f''(x)\) is positive, this means that the function \(f(x)\) has a local minimum at \(x = \sqrt{6}\).

Step 6: Find the corresponding value of y and the minimum sum

Plug the value \(x = \sqrt{6}\) back into the expression for \(y\) in terms of \(x\): \[y = \frac{12}{\sqrt{6}}\] To find the minimum sum, plug the values for \(x\) and \(y\) into the sum \(2x+y\): \[2\sqrt{6} + \frac{12}{\sqrt{6}}\] Therefore, the positive numbers \(x\) and \(y\) satisfying the equation \(x y = 12\) while minimizing the sum \(2x+y\) are \(x = \sqrt{6}\) and \(y = \frac{12}{\sqrt{6}}\), resulting in the minimum sum of \(2\sqrt{6} + \frac{12}{\sqrt{6}}\).