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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. $$f(x, y)=1-2 x^{2}-3 y^{2}$$

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.
See the step by step solution

## 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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