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Problem 1

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. \(f(x, y)=1-2 x^{2}-3 y^{2}\)

Short Answer

Expert verified
The critical point of the function \(f(x, y) = 1-2x^2-3y^2\) is (0, 0), found by setting the partial derivatives \(\frac{\partial f}{\partial x} = -4x\) and \(\frac{\partial f}{\partial y} = -6y\) equal to 0. Using the second derivative test with the Hessian matrix determinant, \(D = 24\), it is determined that (0, 0) is a relative maximum. Therefore, the relative extrema of the function are the relative maximum (0, 0) with a function value of f(0, 0) = 1.
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Step 1: Find the partial derivatives

First, find the partial derivatives with respect to x and y of the given function: \(\frac{\partial f}{\partial x} = -4x\) \(\frac{\partial f}{\partial y} = -6y\)

Step 2: Find the critical points

Set the partial derivatives equal to 0 and solve for x and y: \(-4x = 0\) => \(x = 0\) \(-6y = 0\) => \(y = 0\) The only critical point is (0, 0).

Step 3: Use the second derivative test to classify the critical point

Now we need to find the second partial derivatives of the function: \(\frac{\partial^2 f}{\partial x^2} = -4\) \(\frac{\partial^2 f}{\partial y^2} = -6\) \(\frac{\partial^2 f}{\partial x \partial y} = 0\) \(\frac{\partial^2 f}{\partial y \partial x} = 0\) Next, find the determinant of the Hessian matrix: \(D = \begin{vmatrix} \frac{\partial^2 f}{\partial x^2}&\frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial y \partial x}&\frac{\partial^2 f}{\partial y^2} \end{vmatrix} = (-4)(-6) - (0)(0) = 24\) Since \(D > 0\) and \(\frac{\partial^2 f}{\partial x^2} < 0\), the critical point (0, 0) is a relative maximum.

Step 4: Determine the relative extrema

Since the function has only one critical point, which is a relative maximum, the relative extrema of the function are: Relative Maximum: (0, 0) And the value of the function at this point is: \(f(0, 0) = 1 - 2(0)^2 - 3(0)^2 = 1\)

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