Problem 12 - Calculus Early Transcendentals

Question

Find numbers \(x\) and \(y\) satisfying the equation \(3 x+y=12\) such that the product of \(x\) and \(y\) is as large as possible.

Step-by-Step Solution

Answer

Answer: The maximum product of x and y occurs when x = 2 and y = 6.

Step 1: Express y in terms of x

From the given equation \(3x+y=12\), we can express \(y\) in terms of \(x\) as follows: \(y=12-3x\)

Step 2: Create the function representing the product of x and y

Now that we have \(y\) in terms of \(x\), we can create the function \(P(x)\) representing the product of \(x\) and \(y\): \(P(x)=x(12-3x)\)

Step 3: Find the maximum of P(x) using calculus

First, we need to find the derivative of the function \(P(x)\) with respect to \(x\): \(P'(x)=\frac{d}{dx}(x(12-3x))\) Using the product rule, we have: \(P'(x)=(1)(12-3x) + (x)(-3)\) Simplifying: \(P'(x)=12-3x-3x\) \(P'(x)=12-6x\) Now let's find the critical points of \(P(x)\) by setting its derivative equal to zero: \(P'(x)=0 \Rightarrow 12-6x=0\) Solving for \(x\), we get: \(x=2\) Now let's analyze the second derivative to determine if this critical point is a maximum or minimum: \(P''(x)=\frac{d^2}{dx^2}P(x) = \frac{d^2}{dx^2}(12-6x)\) \(P''(x)=-6\) Since \(P''(x)\) is negative, we can conclude that at \(x=2\), the function \(P(x)\) has a maximum point.

Step 4: Find the value of y corresponding to x=2

Now, we know that the maximum product occurs when \(x=2\). Let's find the corresponding value for \(y\) using the equation we derived in step 1: \(y=12-3x\) Substituting \(x=2\), we get: \(y=12-3(2)=12-6=6\)

Step 5: Final Answer

The maximum possible product of \(x\) and \(y\) occurs when \(x=2\) and \(y=6\), in which case their product is \(2\times6=12\).