Question

Divide 8 into two parts \(x, y\), such that \(x y(x-y)\) is a maximum. (Note that this was posed in the days before calculus.)

Short answer

Answer: The maximum value for the expression \(xy(x-y)\) occurs when \(x = \frac{8}{3}\) and \(y = \frac{16}{3}\).

Step-by-step solution

Express y in terms of x

Since \(x + y = 8\), we can write y in terms of x: $$y = 8 - x$$

Substitute y into the expression

Now we can substitute this expression for y into the given expression \(xy(x-y)\): $$x(8-x)(x-(8-x))$$

Simplify the expression

Simplify the expression: $$x(8-x)(x-8+2x)$$ Expand the expression: $$x(8-x)(3x-8)$$

Find critical points by examining the zeros

To find the critical points, we'll need to examine when the expression is equal to zero. Notice that our expression can be written as a product of three terms: x, \((8-x)\), and \((3x - 8)\). This product will be equal to zero when any of the factors is equal to zero: $$x=0,\ x=8,\ x=\frac{8}{3}$$

Analyze the critical points

There are three critical points, x=0, x=8, and \(x=\frac{8}{3}\). Since we are dividing 8 into two parts x and y, x=0 and x=8 do not make sense because they would give one part as 0 and the other as 8. So, we can discard these two critical points. Now let's check \(x=\frac{8}{3}\). Since y = 8 - x, this gives \(y = 8 - \frac{8}{3} = \frac{16}{3}\). So, we have $$x = \frac{8}{3}$$ and $$y = \frac{16}{3}$$.

Verify the maximum value

We can verify that the expression gives the maximum value by checking the first derivative. But since we can't use calculus, we can check the second derivative by verifying the signs around the critical point x=\(\frac{8}{3}\) (by plug in some numbers close to x) to see if they change from positive to negative. You can check by plugging test values(e.g. x=2 and x=3) into the expression which results in positive and negative values, respectively, indicating a maximum value at x=\(\frac{8}{3}\). Therefore, the maximum value for the expression \(xy(x-y)\) occurs when \(x = \frac{8}{3}\) and \(y = \frac{16}{3}\).

Key concepts

Extreme Value Problem

An extreme value problem in mathematics is a situation where you need to find the maximum or minimum value of a function under certain constraints. When you're not using calculus, finding these extreme values requires an understanding of the behavior of the algebraic expression. In our textbook exercise, the task was to divide the number 8 into two parts, labeled as variables x and y, such that the product of x, y, and the difference between x and y (yz))^2(2k_{scr})$, quadratic equations become much simpler to handle. Remember, simplifying algebraic expressions is not only about mechanical manipulations but understanding the structure of equations to solve problems effectively.

Critical Points Analysis

In the absence of calculus tools, the analysis of critical points is typically done by finding values of variables that make the derivative of the function equal to zero or by considering where the function is undefined. However, without calculus, you look for the values that make the expression equal to zero. These values are potential candidates for local maxima or minima.

In the exercise, the critical points were found by setting each factor of the product equal to zero. Since some critical points can lead to a maximum or minimum value, it’s important to check each one within the domain of the problem. We dismissed the critical points where x was 0 or 8 because they didn’t make sense in the context of the problem - dividing eight into two non-zero parts. Analysis of critical points is more than just solving for zeros; it involves understanding the context and domain of the problem to find meaningful solutions.

Quadratic Functions

A quadratic function is a second-degree polynomial function of the form )