Question

Find two positive numbers that satisfy the given requirements. The product is 192 and the sum of the first plus three times the second is a minimum.

Short answer

The two numbers that satisfy the condition are \(x = 24\) and \(y = 8\).

Step-by-step solution

Define the Variables

Since there are two numbers, we can define them as \(x\) and \(y\). Given that their product is 192, we can represent this as \(x.y = 192\). From this equation, we can express \(y\) in terms of \(x\) as \(y = 192 / x\). Next, we are told that the sum of the first number and three times the second number is a minimum which gives us another equation \(S = x + 3y\). We will be minimizing this equation.

Substitute \(y\) into the Sum Equation

Replace \(y\) in the equation \(S = x + 3y\) with \(y = 192 / x\). This results in \(S = x + 3 (192 / x) = x + 576 / x\).

Differentiate S

Taking the derivative of \(S\) with respect to \(x\) (denoted as \(S'\)): \(S' = 1 - 576 / x^2\). This equation gives the slope of the function at any x-value.

Find Critical Points

To find the minimum value of S, set \(S' = 0\) and solve for x. Solving \(1 - 576 / x^2 = 0\) gives two x-values: 24 and -24. Since we're looking for positive numbers, we eliminate -24, leaving x=24.

Substitute \(x\) into \(y\)

Finally, substitute \(x = 24\) into \(y = 192 / x\). This gives \(y = 192 / 24 = 8\).

Verify Solution

The numbers \(x=24\) and \(y=8\) satisfy both given requirements: \(24*8=192\) and the sum \(24 + 3*8 = 36\) is minimized.

Key concepts

Critical Points in Calculus

In calculus, critical points are essential when analyzing the behavior of functions. They are points on the function where the derivative is either zero or undefined. To find them, one typically takes the derivative of the function and sets it equal to zero, then solves for the variable.

For example, consider the problem where we want to minimize the sum of a number and three times another number, subject to the constraint that their product is 192. After defining our variables as x and y, we express y in terms of x to formulate a single-variable function that represents the sum. The derivative of this function with respect to x provides the rate of change of the sum. Setting this derivative equal to zero gives us the critical points which potentially are the points of minimum (or maximum) sum.

Derivative Applications

The process of differentiation is not only about finding slopes of tangents but is extensively used in optimization problems. By applying derivatives, we can determine the minimum or maximum values of functions that model real-world scenarios.

Upon differentiating the sum S = x + 3y, and substituting the constraint y = 192 / x, the derivative S' = 1 - 576 / x^2 is obtained. This derivative indicates how the sum changes with x. Where this derivative is zero are where the sum's rate of change switches, hence where we’re likely to find a minimum or maximum. By focusing on these points, we can directly apply this concept to solve for the smallest possible sum given the constraints in the exercise.

Minimization Problems

Minimization problems are a type of optimization problem where we seek the smallest possible value of a function within a given domain. In the textbook exercise, the goal is to minimize the sum S under certain conditions.

In the given problem, once we identify the critical point x = 24, we can determine the corresponding y value and check that they indeed minimize the function by showing that the derivative changes sign at that point. This process often involves a second-derivative test or inspecting the values around the critical point. The solution, where x = 24 and y = 8, demonstrates the smallest possible sum (36), thus successfully solving the minimization problem.