Question

Locate the position of any stationary points of the following functions: $$ f(x, y)=\frac{x^{3}}{3}+3 x^{2}+x y+\frac{y^{2}}{2}+6 y $$

Short answer

Answer: The stationary points of the function are (-6, 0) and (1, -7).

Step-by-step solution

Find the partial derivatives with respect to x and y.

We are given the function $$ f(x, y) = \frac{x^3}{3} + 3x^2 + xy + \frac{y^2}{2} + 6y $$ First, we find the partial derivative with respect to x, denoted as \(\frac{\partial f}{\partial x}\): $$ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left(\frac{x^3}{3} + 3x^2 + xy + \frac{y^2}{2} + 6y\right) = x^2 + 6x + y $$ Next, we find the partial derivative with respect to y, denoted as \(\frac{\partial f}{\partial y}\) $$ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(\frac{x^3}{3} + 3x^2 + xy + \frac{y^2}{2} + 6y\right) = x + y + 6 $$

Set the partial derivatives equal to zero and solve for x and y.

We now have a system of equations with partial derivatives: $$ \begin{cases} x^2 + 6x+y = 0 \\ x + y + 6 = 0 \end{cases} $$ Let's solve this system of equations to find the stationary points. From the second equation, we can isolate y and obtain: $$ y = -x - 6 $$ Now, substitute this expression for y into the first equation to get: $$ x^2 + 6x - x - 6 = 0 $$ Simplifying, we have: $$ x^2 + 5x -6= 0 $$ Factoring this quadratic equation, we find: $$ (x + 6)(x -1)= 0 $$ So the possible values for x are -6 and 1. For x = -6, we have y = 0, and for x = 1, we have y = -7.

Write down the stationary points.

We found the two stationary points of the function by solving the system of equations given by the partial derivatives. Therefore, the stationary points for the function \(f(x, y)\) are located at: $$ (-6, 0) \quad \text{and} \quad (1, -7) $$

Key concepts

Partial Derivatives

Partial derivatives help us understand how a function changes with respect to each variable independently. Given a function of multiple variables, here denoted as \(f(x, y)\), a partial derivative with respect to \(x\) is calculated by differentiating the function while considering \(y\) as a constant.

In simpler terms, think of it as changing one variable while freezing the other, seeing how this change impacts the function. This is analogous to climbing a multivariable surface and checking the slope when moving only horizontally or vertically.

• The partial derivative of \(f\) with respect to \(x\), \(\frac{\partial f}{\partial x}\), tells us the function’s rate of change along the \(x\)-axis.
• Similarly, \(\frac{\partial f}{\partial y}\) reveals the rate of change along the \(y\)-axis.

For the given function, when finding the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\), be mindful to apply the derivative rules to each term carefully while treating the other variable as a constant.

Stationary Points

Stationary points occur where the slope of the tangent to the surface is zero. This means that the function does not increase or decrease at these points; it stays constant. In mathematical terms, stationary points are where the partial derivatives of the function are equal to zero.

To find these points, set the calculated partial derivatives equal to zero. By solving these equations, you unearth places on the function's surface that are flat, akin to finding a plateau or valley on a terrain.

• If \(\frac{\partial f}{\partial x} = 0\) and \(\frac{\partial f}{\partial y} = 0\), a stationary point is present.
• This leads to a system of equations. Solving these will provide the exact coordinates of these flat spots.

In this exercise, the stationary points were found by solving the derived equations: \(x^2 + 6x + y = 0\) and \(x + y + 6 = 0\). This resulted in the solutions \((-6, 0)\) and \((1, -7)\), indicating the function's stationary behavior.

System of Equations

A system of equations is crucial when dealing with functions of several variables to find solutions that satisfy all conditions at once. Here, these equations arise from setting the partial derivatives to zero, aiming to identify stationary points.

A system can involve two or more equations working together. Solving such systems provides values for multiple variables simultaneously, corresponding to the scenarios where all conditions are met.

• To solve, isolate one variable in one equation and substitute into the others.
• This can reduce the problem into a simpler one involving fewer equations.

In this example, we first expressed \(y = -x - 6\) from \(x + y + 6 = 0\). By substituting this back into \(x^2 + 6x + y = 0\), a simpler quadratic equation emerged: \(x^2 + 5x - 6 = 0\). Solving such equations—often using methods like factoring—yields the points of interest for the original multivariable function.