To find the relative extrema of the function \( R(x, y) \), we need to find the critical points by setting the partial derivatives with respect to \( x \) and \( y \) to zero and solving the resulting system of equations. Then, we can use the second derivative test to classify the critical points.
- Compute the partial derivatives of \( R(x, y) \) with respect to \( x \) and \( y \).
- Set the partial derivatives equal to zero to find the critical points.
- Use the second derivative test to determine the nature of each critical point (i.e., whether it is a local maximum, local minimum, or saddle point).
To find the relative extrema of the function \( R(x, y) \), we first need to find the partial derivatives with respect to \( x \) and \( y \).
\[
R(x, y) = -5x^2 - 8y^2 - 2xy + 42x + 102y
\]
The partial derivative with respect to \( x \) is:
\[
\frac{\partial R}{\partial x} = \frac{\partial}{\partial x} (-5x^2 - 8y^2 - 2xy + 42x + 102y)
\]
\[
\frac{\partial R}{\partial x} = -10x - 2y + 42
\]
The partial derivative with respect to \( y \) is:
\[
\frac{\partial R}{\partial y} = \frac{\partial}{\partial y} (-5x^2 - 8y^2 - 2xy + 42x + 102y)
\]
\[
\frac{\partial R}{\partial y} = -16y - 2x + 102
\]
To find the critical points, we set the partial derivatives equal to zero and solve for \( x \) and \( y \).
\[
-10x - 2y + 42 = 0
\]
\[
-16y - 2x + 102 = 0
\]
We solve the system of linear equations:
- \(-10x - 2y + 42 = 0\)
- \(-16y - 2x + 102 = 0\)
First, solve equation 1 for \( y \):
\[
-10x - 2y + 42 = 0
\]
\[
-2y = 10x - 42
\]
\[
y = -5x + 21
\]
Substitute \( y = -5x + 21 \) into equation 2:
\[
-16(-5x + 21) - 2x + 102 = 0
\]
\[
80x - 336 - 2x + 102 = 0
\]
\[
78x - 234 = 0
\]
\[
78x = 234
\]
\[
x = 3
\]
Now substitute \( x = 3 \) back into \( y = -5x + 21 \):
\[
y = -5(3) + 21
\]
\[
y = -15 + 21
\]
\[
y = 6
\]
The critical point is:
\[
\boxed{(x, y) = (3, 6)}
\]
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