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Chapter 21: Problem 4

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

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

Step by step solution

01

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 \(\frac{\partial f}{\partial y}\) $$ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left(\frac{x^3}{3} + 3x^2 + xy + \frac{y^2}{2} + 6y\right) = x + y + 6 $$
02

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.
03

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: $$ (-6, 0) \quad \text{and} \quad (1, -7) $$

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Most popular questions from this chapter

If \(f(x, t)=\mathrm{e}^{2 x}\) find \(f(0.5,3)\).

Find all the second partial derivatives in each of the following cases: (a) \(z=x \sin y(\) b) \(z=y \cos x\) (c) \(z=y \mathrm{e}^{2 x}\left(\right.\) d) \(z=y \mathrm{e}^{-x}\)

If \(z=f(x, y)=\sin (x+y)\) find \(f\left(20^{\circ}, 30^{\circ}\right)\) where the inputs are angles measured in degrees.

Locate the position of any stationary points of the following functions: $$ f(x, y)=4 x y-2 x^{2} y $$

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