Problem 37 - A History of Mathematics: An Introduction

Question

37\. Determine all the relative extrema for \(V=x^{3}+y^{2}-\) \(3 x y+(3 / 2) x\), and for each one determine whether it is a maximum or minimum. Compare your answer with that of Euler.

Step-by-Step Solution

Answer

Answer: The function V has one relative minimum at the point \((1, \frac{3}{2})\).

Step 1: Find the partial derivatives

To find the partial derivatives, we will differentiate V with respect to x and y, keeping the other variable constant. The partial derivatives are given by: $$\frac{\partial V}{\partial x} = 3x^2 - 3y + \frac{3}{2}$$ $$\frac{\partial V}{\partial y} = 2y - 3x$$

Step 2: Find the critical points

To find the critical points, we need to set both partial derivatives equal to zero and solve for x and y: $$3x^2 - 3y + \frac{3}{2} = 0$$ $$2y - 3x = 0$$ Solving this system of equations, we get two critical points: \((x_1, y_1) = (0, 0)\) and \((x_2, y_2) = (1, \frac{3}{2})\).

Step 3: Apply the second partial derivative test

To apply the second partial derivative test, we need to find the second partial derivatives and calculate the determinant of the Hessian matrix. First, let's find the second partial derivatives: $$\frac{\partial^2 V}{\partial x^2} = 6x$$ $$\frac{\partial^2 V}{\partial y^2} = 2$$ $$\frac{\partial^2 V}{\partial x \partial y} = \frac{\partial^2 V}{\partial y \partial x} = -3$$ The determinant of the Hessian matrix is given by: $$D(x,y) = \frac{\partial^2 V}{\partial x^2} \cdot \frac{\partial^2 V}{\partial y^2} - \left(\frac{\partial^2 V}{\partial x \partial y}\right)^2 = (6x)(2) - (-3)^2$$ Now let's evaluate the determinant at both critical points: - At \((x_1, y_1) = (0, 0)\): $$D(0, 0) = (6 \cdot 0)(2) - (-3)^2 = 0 - 9 < 0$$ Since the determinant is negative, \((0, 0)\) is a saddle point, and hence not a relative extrema. - At \((x_2, y_2) = (1, \frac{3}{2})\): $$D(1, \frac{3}{2}) = (6 \cdot 1)(2) - (-3)^2 = 12 - 9 = 3 > 0$$ Since the determinant is positive, we need to check the value of \(\frac{\partial^2 V}{\partial x^2}\) at this point: $$\frac{\partial^2 V}{\partial x^2}(1, \frac{3}{2}) = 6 \cdot 1 = 6 > 0$$ Since \(\frac{\partial^2 V}{\partial x^2} > 0\), the point \((1, \frac{3}{2})\) is a relative minimum of V. In conclusion, we found one relative extrema for the function V: a minimum at \((1, \frac{3}{2})\). This result is consistent with Euler's.

Key Concepts

Partial Derivatives

Partial derivatives are essential in calculus, especially when dealing with functions of multiple variables. In this exercise, the function \(V(x, y) = x^3 + y^2 - 3xy + \frac{3}{2}x\) involves two variables, \(x\) and \(y\). When we compute the partial derivative with respect to \(x\), we treat \(y\) as a constant. Conversely, computing the partial derivative with respect to \(y\) treats \(x\) as a constant.

• For \(\frac{\partial V}{\partial x}\), we differentiate each term that includes \(x\) while keeping \(y\) constant, yielding: \(3x^2 - 3y + \frac{3}{2}\).
• For \(\frac{\partial V}{\partial y}\), we differentiate each term that includes \(y\), resulting in: \(2y - 3x\).

Once partial derivatives are calculated, they help us identify points where the function may attain a relative extremum or saddle point by solving \(\frac{\partial V}{\partial x} = 0\) and \(\frac{\partial V}{\partial y} = 0\). This is a crucial prelude to finding critical points.

Critical Points

Critical points occur where the gradient of the function \(V(x, y)\) is zero or undefined. In this exercise, the critical points are found by solving the system of equations obtained from setting both partial derivatives to zero.

• \(3x^2 - 3y + \frac{3}{2} = 0\)
• \(2y - 3x = 0\)

By solving these equations simultaneously, we obtain the critical points \((0, 0)\) and \((1, \frac{3}{2})\). At these locations, the function \(V\) potentially has local maxima, minima, or saddle points.

Critical points are the gateway to deeper analysis with the Hessian matrix and Second Partial Derivative Test, which help determine their nature.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function of several variables. It gives insight into the local curvature of the function. For the given function \(V(x, y)\), the Hessian matrix, \(H\), is constructed as follows:

\[H = \begin{pmatrix}\frac{\partial^2 V}{\partial x^2} & \frac{\partial^2 V}{\partial x \partial y} \\frac{\partial^2 V}{\partial y \partial x} & \frac{\partial^2 V}{\partial y^2}\end{pmatrix} = \begin{pmatrix}6x & -3 \-3 & 2\end{pmatrix} \]
The Hessian matrix is primarily used in the Second Partial Derivative Test to determine the nature (maximum, minimum, or saddle point) of each critical point.

Second Partial Derivative Test

The Second Partial Derivative Test uses the Hessian matrix to classify critical points. For each critical point, the determinant \(D(x, y)\) of the Hessian matrix is calculated. This determinant helps decide the nature of the critical point:

• If \(D(x, y) > 0\) and \(\frac{\partial^2 V}{\partial x^2} > 0\), the function has a local minimum.
• If \(D(x, y) > 0\) and \(\frac{\partial^2 V}{\partial x^2} < 0\), the function has a local maximum.
• If \(D(x, y) < 0\), the point is a saddle point, indicating neither a local maximum nor minimum.

Evaluating the determinant at the critical points of \(V(x, y)\) found that \((0, 0)\) is a saddle point, as \(D(0,0) < 0\). At \((1, \frac{3}{2})\), the positive determinant combined with \(\frac{\partial^2 V}{\partial x^2} > 0\) confirms a local minimum, aligning with Euler's original findings.