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Chapter 4: Problem 13

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

Short Answer

Expert verified
Answer: The positive numbers x and y that satisfy the equation xy = 12 and minimize the sum 2x + y are x = 鈭�6 and y = 2鈭�6.

Step by step solution

01

Write an equation for y in terms of x from the given equation

The given equation is \(xy=12\). We can solve it for \(y\) to get: \[y = \frac{12}{x}\]
02

Substitute the equation for y in terms of x into the sum formula

The sum formula is \(2x+y\). Substitute the equation for \(y\) from Step 1 into the sum formula: \[2x + \frac{12}{x}\]
03

Find the derivative of the sum function with respect to x

The function to minimize is \(f(x) = 2x + \frac{12}{x}\). Find the derivative of the function with respect to x: \[f'(x) = \frac{d}{dx} \left(2x + \frac{12}{x}\right)\] Using the power rule, we have: \[f'(x) = 2 - \frac{12}{x^2}\]
04

Set the derivative equal to zero and solve for x

To find the critical points of the function, we set the derivative equal to zero and solve for x: \[2 - \frac{12}{x^2} = 0\] Multiply both sides of the equation by \(x^2\), we get: \[2x^2 - 12 = 0\] Now, we can solve for \(x\). Divide both sides by \(2\): \[x^2 = 6\] Find the square root of both sides: \[x = \sqrt{6}\] Since we are only looking for positive solutions, we only consider the positive square root of 6.
05

Find the corresponding y-value

Now that we have found the value of \(x\), we can find the value of \(y\). Recall that \(y=\frac{12}{x}\), and plug in the value of \(x=\sqrt{6}\): \[y = \frac{12}{\sqrt{6}}\] To simplify this, multiply the numerator and the denominator by \(\sqrt{6}\): \[y = \frac{12\sqrt{6}}{6}\] Divide by 6: \[y = 2\sqrt{6}\]
06

State the solution

The positive numbers \(x\) and \(y\) that satisfy the equation \(xy=12\) and minimize the sum \(2x+y\) are: \[x=\sqrt{6}\] and \[y=2\sqrt{6}\]

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