Find the maxima, minima, and saddle points of the function f(x,y) = x2 + xy + 3x + 2y + 5.

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ScanSolve AI Tutor · Accepted Answer

To find the maxima, minima, and saddle points of the function $f(x,y) = x^2 + xy + 3x + 2y + 5$ , we follow these steps:

Find the first partial derivatives and set them to zero to find critical points.
The given function is $f(x,y) = x^2 + xy + 3x + 2y + 5$ .
The partial derivative with respect to $x$ is:
$f_x = \frac{\partial}{\partial x}(x^2 + xy + 3x + 2y + 5) = 2x + y + 3$
The partial derivative with respect to $y$ is:
$f_y = \frac{\partial}{\partial y}(x^2 + xy + 3x + 2y + 5) = x + 2$
Set $f_x = 0$ and $f_y = 0$ :
1. $2x + y + 3 = 0$
2. $x + 2 = 0$
From equation (2):
$x = -2$
Substitute $x = -2$ into equation (1):
$2(-2) + y + 3 = 0$ $-4 + y + 3 = 0$ $y - 1 = 0$ $y = 1$
The only critical point is $(-2,1)$ .
Find the second partial derivatives.
$f_{xx} = \frac{\partial}{\partial x}(2x + y + 3) = 2$ $f_{yy} = \frac{\partial}{\partial y}(x + 2) = 0$ $f_{xy} = \frac{\partial}{\partial y}(2x + y + 3) = 1$
Calculate the discriminant $D(x,y) = f_{xx}f_{yy} - (f_{xy})^2$ .
$D(x,y) = (2)(0) - (1)^2 = 0 - 1 = -1$
Apply the Second Derivative Test to the critical point.
- At $(-2,1)$ : $D(-2,1) = -1$ Since $D(-2,1) < 0$ , the critical point $(-2,1)$ is a saddle point.
The final answer is $\text{Saddle point at } (-2,1)$ .

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