Problem 7
Question
Examine the function for relative extrema and saddle points. $$ f(x, y)=2 x^{2}+2 x y+y^{2}+2 x-3 $$
Step-by-Step Solution
Verified Answer
The function \(f(x, y)=2 x^{2}+2 x y+y^{2}+2 x-3\) has a relative minimum at point (1, -1).
1Step 1 - Find the first partial derivatives
They are, \(f_x = 4x + 2y + 2\) and \(f_y = 2x + 2y\).
2Step 2 - Solve \(f_x = 0\) and \(f_y = 0\)
From \(4x + 2y + 2 = 0\), we get \(2x + y = -1\) and from \(2x + 2y = 0\), we get \(x + y = 0\). From solving the system of equations \(2x + y = -1\) and \(x + y = 0\), we get \(x=1, y=-1\).
3Step 3 - Calculate the second order partial derivatives
The second order partial derivatives are \(f_{xx} = 4\), \(f_{xy}=f_{yx}=2\), and \(f_{yy}=2\).
4Step 4 - Calculate the determinant of the Hessian matrix and analyze critical points
The Hessian matrix, H, is given by \(H = [[f_{xx}, f_{xy}], [f_{yx}, f_{yy}]] = [[4, 2], [2, 2]]\). The determinant of H, denoted as D, is calculated by \(D = f_{xx}*f_{yy} - (f_{xy})^2 = 4*2 - (2)^2 = 4\). Since \(D>0\) and \(f_{xx}>0\), the critical point (1, -1) is a relative minimum.
Other exercises in this chapter
Problem 7
Evaluate the partial integral. $$ \int_{1}^{e} \frac{y \ln x}{x} d x $$
View solution Problem 7
Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \text { Maximize } f(x, y)=2 x+2 x y+y \quad 2 x
View solution Problem 7
Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ f(x, y)=\sqrt{x^{2}+y^{2}} $$
View solution Problem 7
Find the function values. \(V(r, h)=\pi r^{2} h\) (a) \(V(3,10)\) (b) \(V(5,2)\)
View solution