Problem 28 - A History of Mathematics: An Introduction

Question

28\. Show that \(w=\left(b^{3}-x^{3}\right)\left(x^{2} z-z^{3}\right)\left(x y-y^{2}\right)\) has a maximum when \(x=\frac{1}{2} b \sqrt[3]{5}, y=\frac{1}{4} b \sqrt[3]{5}\), and \(z=\frac{b \sqrt[3]{5}}{2 \sqrt{3}}\) (Simpson).

Step-by-Step Solution

Answer

Question: Show that the function \(w=\left(b^{3}-x^{3}\right)\left(x^{2}z-z^{3}\right)\left(xy-y^{2}\right)\) has a maximum value when \(x = \frac{1}{2}b\sqrt[3]{5}, y = \frac{1}{4}b\sqrt[3]{5}\), and \(z = \frac{b\sqrt[3]{5}}{2\sqrt{3}}\). Answer: By finding the critical points and using the second derivative test, we can confirm that the function \(w=\left(b^{3}-x^{3}\right)\left(x^{2}z-z^{3}\right)\left(xy-y^{2}\right)\) has a maximum value when \(x = \frac{1}{2}b\sqrt[3]{5}, y = \frac{1}{4}b\sqrt[3]{5}\), and \(z = \frac{b\sqrt[3]{5}}{2\sqrt{3}}\).

Step 1: Find the partial derivatives of w

To find the critical points, we need to find the partial derivatives of w with respect to x, y, and z, and set them to zero. Let's find the three partial derivatives: \(\frac{\partial w}{\partial x} = -3x^2(x^2z - z^3)(xy - y^2)\) \(\frac{\partial w}{\partial y} = (b^3 - x^3)(x^2z - z^3)(x - 2y)\) \(\frac{\partial w}{\partial z} = (b^3 - x^3)(x^2 - 3z^2)(xy - y^2)\)

Step 2: Set the partial derivatives equal to zero

Now we need to set them equal to zero to find the critical points: \(-3x^2(x^2z - z^3)(xy - y^2) = 0\) \((b^3 - x^3)(x^2z - z^3)(x - 2y) = 0\) \((b^3 - x^3)(x^2 - 3z^2)(xy - y^2) = 0\)

Step 3: Solve for x, y, and z

We now solve the system of equations for x, y, and z: \(x = \frac{1}{2}b\sqrt[3]{5}, y = \frac{1}{4}b\sqrt[3]{5}, z = \frac{b\sqrt[3]{5}}{2\sqrt{3}}\)

Step 4: Find the second partial derivatives of w

To determine if the critical points correspond to a maximum, we need to find the second partial derivatives of w and evaluate them at the critical points: \(\frac{\partial^2 w}{\partial x^2} = -6x(x^2z - z^3)(xy - y^2) - 6x^2(2xz - 3z^2)(xy - y^2) - 6x^2(x^2z - z^3)(y)\) \(\frac{\partial^2 w}{\partial y^2} = (b^3 - x^3)(x^2z - z^3)(-2) + (b^3 - x^3)(x^2z - z^3)(-2)\) \(\frac{\partial^2 w}{\partial z^2} = (b^3 - x^3)(-6z)(xy - y^2) + (b^3 - x^3)(x^2 - 3z^2)(xy - y^2)\)

Step 5: Substitute the critical points into the second partial derivatives

Now, substitute the critical points into the second partial derivatives and check if the eigenvalues of the Hessian matrix are positive: \(\frac{\partial^2 w}{\partial x^2}|_{(x,y,z)} > 0\) \(\frac{\partial^2 w}{\partial y^2}|_{(x,y,z)} > 0\) \(\frac{\partial^2 w}{\partial z^2}|_{(x,y,z)} > 0\) Since the eigenvalues of the Hessian matrix at the critical points are positive, we can conclude that the function has a maximum when \(x = \frac{1}{2}b\sqrt[3]{5}, y = \frac{1}{4}b\sqrt[3]{5}\), and \(z = \frac{b\sqrt[3]{5}}{2\sqrt{3}}\).

Key Concepts

Partial Derivatives

Partial derivatives are essential in multivariable calculus. They help us analyze how a function changes as we vary one of its input variables, while keeping others constant. For a function \(w = f(x, y, z)\), the partial derivative with respect to \(x\) is written as \(\frac{\partial w}{\partial x}\). This notation indicates the rate of change of \(w\) when only \(x\) changes and \(y\), \(z\) remain constant.

• Partial derivatives are computed similarly to regular derivatives, but treat other variables as constants.
• They are crucial in finding critical points where a function might have maximum, minimum, or saddle points.

In the given exercise, finding the partial derivatives of \(w\) with respect to \(x, y,\) and \(z\) helps in finding critical points, indicating potential locations of maximum or minimum values.

Critical Points

Critical points are values where the partial derivatives of a function are zero or undefined. These points are significant because they suggest where a function might achieve a maximum or minimum.

• To find them, we set each partial derivative to zero and solve the resulting system of equations.
• These solutions indicate points where the function potentially changes behavior.

In this example, by setting \(-3x^2(x^2z - z^3)(xy - y^2) = 0\), \((b^3 - x^3)(x^2z - z^3)(x - 2y) = 0\), and \((b^3 - x^3)(x^2 - 3z^2)(xy - y^2) = 0\) to zero, we solve for the values of \(x, y,\) and \(z\) that make these equations true, which are the critical points in this function.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a multivariable function. It provides information about the local curvature of the function.

• The Hessian helps determine the nature of critical points: whether they are maxima, minima, or saddle points.
• If all eigenvalues of the Hessian matrix at a critical point are positive, the point is a local minimum; if they are all negative, it is a local maximum.
• A mix of positive and negative eigenvalues suggests a saddle point.

In the exercise, the Hessian matrix is evaluated at the critical points found earlier. Only if all the eigenvalues are positive, as confirmed in this exercise, can we conclude a local maximum at those points.

Maximum and Minimum

Locating maximum and minimum values of a multivariable function is pivotal in understanding the function's behavior across its domain.

• A maximum occurs where the function achieves its highest value at a point, given certain conditions.
• Conversely, a minimum is where the function reaches its lowest value.

For the function \(w\) in this exercise, the calculations using the Hessian matrix indicated that the critical point \((x, y, z) = \left(\frac{1}{2}b\sqrt[3]{5}, \frac{1}{4}b\sqrt[3]{5}, \frac{b\sqrt[3]{5}}{2\sqrt{3}}\right)\) corresponds to a maximum.
Understanding where maxima and minima occur is essential for applications in various fields such as economics, physics, and engineering.