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Chapter 13: Problem 28

Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=x^{3}+y^{3}-3 x^{2}+6 y^{2}+3 x+12 y+7 $$

Short Answer

Expert verified
The function has a local maximum at the point (1, -2) and a saddle point at the point (-1, -2). No critical points were found where the Second-Partials Test fails.

Step by step solution

01

Calculating the partial derivatives

Start by computing the first order partial derivatives \(f_x\) and \(f_y\) of the function. \(f_x = 3x^2 - 6x + 3\) and \(f_y = 3y^2 + 12y + 12\).
02

Finding the critical points

Set the first order partial derivatives equal to zero and solve the equations to find the critical points. For \(f_x = 0\), we get \(x = 1, -1\). For \(f_y = 0\), we get \(y = -2, -2\). Therefore, the critical points are \((1,-2)\) and \((-1,-2)\).
03

Applying the Second-Partials Test

Compute the second order partial derivatives: \(f_{xx} = 6x - 6\), \(f_{yy} = 6y + 12\), and \(f_{xy} = 0\). Then, calculate the determinant D of the Hessian matrix \(D = f_{xx}*f_{yy} - f_{xy}^2\). Substitute the critical points into D. If D>0 and \(f_{xx}>0\), the point is a local minimum. If D>0 and \(f_{xx}<0\), the point is a local maximum. If D<0, the point is a saddle point. If D=0, the test is inconclusive.
04

Determination of relative extrema and critical points

For the point (1, -2), calculating \(D = -6*(-6) - (0)^2 = 36 > 0\) and \(f_{xx} = -6 < 0\), so it is a local maximum. For the point (-1, -2), calculating \(D = 6*(-6) - (0)^2 = -36 < 0\), so it is a saddle point. Thus, the function has one local maximum at (1,-2) and one saddle point at (-1,-2). The Second-Partials Test does not fail for these critical points.

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