Maximizing the Product of Two Variables under a Constraint in Calculus

Calculus is a branch of mathematics that deals with continuous change. One of the most fundamental applications of calculus is in optimization problems. In this article, we will explore how to find the maximum value of the product xy subject to the constraint x^2 y^2 2000.

Problem Statement and Initial Approach

Given the constraint x^2 y^2 2000, we want to find the maximum value of the product xy. To start, we can express y in terms of x using the constraint:

y sqrt{2000 - x^2}

Substituting this into the product xy, we get:

xy x * sqrt{2000 - x^2}

To find the maximum value of this product, we need to take the derivative of the function fx x * sqrt{2000 - x^2} with respect to x and set it equal to zero.

Derivative Calculation

Let's denote the function we want to maximize as:

fx x * sqrt{2000 - x^2}

Using the product rule, the derivative of fx with respect to x is:

fx' sqrt{2000 - x^2} x * (-x) / (2 * sqrt{2000 - x^2})

Setting this derivative equal to zero to find the critical points:

sqrt{2000 - x^2} - x^2 / (2 * sqrt{2000 - x^2}) 0

By solving the above equation, we get:

x^2 1000

x sqrt{1000} 10sqrt{10}

Substituting x 10sqrt{10} back into the equation y sqrt{2000 - x^2} gives:

y sqrt{1000} 10sqrt{10}

This gives us x y, which is not consistent with xy. Therefore, we need to check integer values of x close to 10sqrt{10} to find a valid pair (x, y).

Checking Integer Values

We will test integer values of x starting from 31 downwards to find a pair that satisfies both the product and the constraint. Here are the values:

Checking x 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4

After evaluating these values, we find that the pair (x, y) (20, 40) satisfies the equation:

x^2 y^2 20^2 40^2 400 1600 2000

Therefore, the maximum value of xy is:

xy 20 * 40 800

Additionally, we find that the pair (x, y) (8, 44) also satisfies the equation:

x^2 y^2 8^2 44^2 64 1936 2000

For the pair (x, y) (8, 44), the value of xy is:

xy 8 * 44 352

Therefore, the maximum value of xy is 800.

Conclusion

The maximum value of xy under the constraint x^2 y^2 2000 is found to be 800 with the pair (x, y) (20, 40). This solution demonstrates the power of calculus in solving optimization problems under constraints.