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Chapter 4: Problem 10

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 numbers x = 2 and y = 6 satisfy the equation and make the product xy as large as possible.

Step by step solution

01

Solve the equation for y

Given the equation 3x + y = 12, we need to solve for y in terms of x: \(y = 12 - 3x\)
02

Write the product xy as a function of x

Now that we have y in terms of x, we can write the product xy as a function of x: \(P(x) = x(12 - 3x) = 12x - 3x^2\)
03

Find the critical points of P(x)

Now we will differentiate P(x) with respect to x and set the derivative equal to 0 to find the critical points: \(P'(x) = \frac{d}{dx}(12x - 3x^2) = 12 - 6x\) Setting \(P'(x)\) equal to 0, we get: \(12 - 6x = 0\) \(6x = 12\) \(x = 2\)
04

Determine if the critical point is a maximum

To determine if the critical point is a maximum, we can use the second derivative test. Calculate the second derivative of P(x): \(P''(x) = \frac{d^2}{dx^2}(12x - 3x^2) = -6\) Since \(P''(x)=-6<0\), the critical point x = 2 corresponds to a maximum of the product P(x).
05

Find the corresponding value of y

Now that we have the value of x = 2 that maximizes the product, we can find the corresponding value of y by substituting x = 2 back into the equation for y: \(y = 12 - 3(2) = 12 - 6 = 6\)
06

Conclusion

The numbers x = 2 and y = 6 satisfy the equation 3x + y = 12 and make the product xy as large as possible.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Points
In calculus, critical points are values in the domain of a function where the derivative is zero or undefined. These points are essential when analyzing the behavior of functions, especially in optimization problems like maximizing or minimizing functions.
To find critical points:
  • Find the derivative of the function.
  • Set the derivative equal to zero, then solve for the variable.
In our product maximization problem, the function is the product of two numbers, rewritten in terms of one variable: \[ P(x) = x(12 - 3x) = 12x - 3x^2 \]By differentiating and setting the derivative to zero, we locate the critical point, which is essential for determining potential maximum or minimum values of the product.
Second Derivative Test
The second derivative test is a method used to determine whether a critical point of a differentiable function is a local maximum, minimum, or a saddle point. This test requires:
  • Calculating the second derivative of the function.
  • Evaluating the second derivative at the critical point(s).
For the function in question, the second derivative is:\[ P''(x) = -6 \]This result is constant and negative. Therefore, when evaluating it at the critical point \(x = 2\), we confirm that the point is a local maximum, as the negative value indicates the curve is concave down at that point.
Product Maximization
Product maximization involves finding the maximum value for the product of two quantities under given constraints. In this exercise, the product we're examining is \( xy \) subject to the equation:\[ 3x + y = 12 \]To maximize this product:
  • Express one variable in terms of the other using the constraint equation.
  • Substitute this expression into the product, getting it as a function of a single variable.
By applying these steps, we obtained \( P(x) = 12x - 3x^2 \), which allowed us to use calculus tools like derivatives to find the optimal values for \(x\) and \(y\) that maximize the product.
Differentiation
Differentiation, a core concept in calculus, allows us to find the rate at which a function is changing at any point. It's an essential tool for solving optimization problems as shown in this example.
To differentiate a function effectively:
  • Apply differentiation rules such as the power rule, product rule, and chain rule.
  • Calculate the derivative at points of interest, particularly critical points.
In our problem, we used differentiation to determine:\[P'(x) = \frac{d}{dx}(12x - 3x^2) = 12 - 6x\]By solving \(P'(x)=0\), we established the critical point \(x=2\), leading us to optimize the product \(xy\), thus illustrating the practical power of differentiation.

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