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Chapter 4: Problem 12

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.

Short Answer

Expert verified
Answer: The values of x and y that satisfy the equation and maximize their product are x = 2 and y = 6.

Step by step solution

01

Write the given equation and constraints

We are given the equation \(3x + y = 12\). Our goal is to find the values of \(x\) and \(y\) such that their product, \(P = xy\), is as large as possible.
02

Solve the given equation for one variable

Let's solve the given equation for one variable; we can choose either \(x\) or \(y\). For this solution, we will solve for \(y\): $$y = 12 - 3x$$
03

Replace the variable y with its expression in the product P

Now that we have an expression for \(y\) in terms of \(x\), we can replace it in the product \(P = xy\): $$P = x(12 - 3x) = 12x - 3x^2$$
04

Find critical points by taking the derivative of P with respect to x

To maximize the product P, we need to find the critical points. We take the derivative of P with respect to x and set it equal to 0: $$\frac{dP}{dx} = 12 - 6x$$ Now we set the derivative equal to 0 and solve for \(x\): $$12 - 6x = 0$$ $$x = 2$$
05

Find the value of y corresponding to the critical point x=2

Now that we have the value of \(x\) that maximizes the product, we can find the corresponding value of \(y\) using the expression for \(y\) in terms of \(x\): $$y = 12 - 3(2) = 12 - 6 = 6$$
06

State the solution

The values of \(x\) and \(y\) satisfying the equation \(3x + y = 12\) that maximize the product \(P = xy\) are \(x = 2\) and \(y = 6\). The maximum product is \(P = xy = 12\).

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