Unlocking the Minimum Value of Complex Functions: A Vector Calculus Approach

Understanding the smallest value that a mathematical function can take is a cornerstone of calculus and has vast applications in fields from engineering to data science. This article explores how to find the minimum value of the function u(x, y) 4x^2y^2 - x^2y - xy 3x^2 5y^2 - 4x - 5y 9 and a similar function using vector calculus. We'll cover the step-by-step process and discuss the significance of such calculations.

Existence of the Minimum Value

Given the function f(x, y) 4x^2y^2 - x^2y - xy 3x^2 5y^2 - 4x - 5y 9, we seek to determine its minimum value. We approach this problem using techniques from vector calculus, including vector fields, gradients, and Hessians. The key steps are as follows:

Determine the critical points where the gradient of the function is zero. Classify these points as local or global minima using the second derivative test. Evaluate the function at the identified minimum points to find the smallest value.

Step-by-Step Solution

Let's consider the function f(x, y) 4x^2y^2 - x^2y - xy 3x^2 5y^2 - 4x - 5y 9 first. To find its critical points, we need to solve the following system of equations derived from the gradient:

[ abla f(x, y) begin{pmatrix}0 0 end{pmatrix}]

This involves computing the partial derivatives and setting them equal to zero:

[ begin{cases} frac{partial f}{partial x} 8xy^2 - 2xy - 1 0 frac{partial f}{partial y} 8yx^2 - 2yx - 1 0 end{cases} ]

Solving for Critical Points

We can simplify and solve the above system to find the critical points. First, let's multiply the first equation by x and the second by y, then subtract the latter from the former:

[2x^2 - y^2 - x - y 0]

By solving for x, we get:

[x y quad text{or} quad x -frac{1}{2} - y]

For the case where x -frac{1}{2} - y, substituting back into the gradient equations:

[8(-frac{1}{2} - y)^2y^2 (frac{1}{2} y)y (frac{1}{2} y) - 1 0]

This simplifies to a quadratic equation in y, which typically has no real solutions. However, for x y, we substitute and find:

[8y^3 - 1 0 quad Rightarrow quad y -frac{1}{2}]

Hence, the critical point is (x, y) (-frac{1}{2}, -frac{1}{2}).

Classifying the Critical Point

To determine if this critical point is a local minimum, we analyze the second derivatives (Hessian matrix) and its determinant:

[det H_f(x, y) 120]

Since the determinant is positive and the second partial derivative with respect to x is positive, x -frac{1}{2}, y -frac{1}{2} is indeed a local minimum. Evaluating the function at this point:

[f(-frac{1}{2}, -frac{1}{2}) frac{1}{4}]

Therefore, the minimum value of f(x, y) is (frac{1}{4}).

General Function Analysis

Now consider the more complex function:

u(x, y) 4x^2y^2 - x^2y - xy 3x^2 5y^2 - 4x - 5y 9

Our friend believes that this problem involves an ellipse or a hyperbola. To understand why, let's transform the function into polar coordinates:

[x rcos theta, quad y rsin theta]

Substituting these into u(x, y), we get:

[u(r, theta) 4r^4cos^2thetasin^2theta - r^3cos^2thetasintheta - r^2costhetasintheta 3r^2cos^2theta 5r^2sin^2theta - 4rcostheta - 5rsintheta 9]

Given that θ ranges from 0 to π/2, we need to minimize u(r, θ) over all possible values of r and θ.

The function u(r, theta) can be seen as a combination of polynomial terms in r and trigonometric functions of θ. The presence of trigonometric terms suggests that the function's behavior changes with the angle θ, and thus, the minimum could occur at different angles. However, the term 4r^4cos^2thetasin^2theta and the cross-terms involving costhetasintheta suggest the function's complexity, which could indeed involve an ellipse or hyperbola depending on the specific values of theta.

Given the constraints, the minimum value of u(r, theta) can be found by evaluating the function at critical points derived from the gradient equations, similar to the earlier approach.

Conclusion

This article has explored the method for finding the minimum value of complex functions using vector calculus. By applying the principles of vector fields, gradients, and Hessians, we can determine the smallest value that a given function can take. The example provided showcases how to approach optimization problems, and the general function analysis illustrates the importance of transforming functions into different coordinate systems for a clearer understanding.

The findings can be summarized as follows: the minimum value of u(x, y) is ..., and by transforming into polar coordinates, the problem's behavior can be better understood, potentially involving ellipses or hyperbolas.