Optimizing the Product of x and y Given the Equation 2xy10

In mathematics, optimizing the product of two variables given an equation can provide valuable insights into the relationship between these variables. In this article, we will explore the process of finding the maximum value of the product xy when provided with the equation 2xy10.

Introduction to the Problem

The equation 2xy10 can be simplified to express one variable in terms of the other. By doing so, we can then manipulate the expression to find the product xy.

Solving for y in Terms of x

Given 2xy10, we can isolate y as follows:

Step 1: Express y in terms of x

[2xy 10 implies y frac{10}{2x} frac{5}{x}]

Step 2: Substitute y into the product xy

Substituting y frac{5}{x} into the product xy gives:

[xy x cdot frac{5}{x} 5]

In this case, the product xy 5, which is a constant and does not depend on the values of x and y. This shows that the product xy is maximized (or minimized, since it's constant) when y is expressed in terms of x as shown above.

Exploring the Optimized Product Using Calculus

While the product is a constant in this case, exploring the process of optimization can help us understand the underlying principles. We will use calculus to find the extremum of the product xy under the constraint 2xy10.

Step 1: Formulate the function to be optimized

The function to be optimized is:

[f(x, y) xy frac{10}{2} - frac{2y^2}{2} 5 - y^2]

Step 2: Express y in terms of x

As derived earlier, y frac{5}{x}. Substituting this into the expression for xy gives:

[xy x cdot frac{5}{x} 5]

Step 3: Use Calculus to find the maximum value

Define the function f(y) frac{5}{2}y - frac{y^2}{2}. To find the critical points, we take the derivative and set it to zero:

[f'(y) frac{5}{2} - y 0 implies y frac{5}{2} 2.5]

Next, we evaluate the second derivative to determine the nature of this critical point:

[f''(y) -1]

Since the second derivative is negative, this indicates a maximum. Therefore, the maximum value of xy is:

[xy left(frac{5}{2}right) cdot left(frac{5}{2} cdot 2right) frac{25}{4} 12.5]

This shows that the maximum value of the product xy given the constraint 2xy10 is 12.5.

Conclusion

In conclusion, the product xy given the equation 2xy10 can be maximized using both algebraic and calculus methods. While the algebraic approach reveals a constant product of 5, the calculus approach provides a deeper understanding of the behavior of the function under the given constraint. The maximum value of the product xy is 12.5 when y frac{5}{2} and x frac{5}{2}.

By understanding these techniques, one can apply similar methods to optimize other expressions and solve a wide range of mathematical and real-world problems.