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Find all critical points of the following function. \( f(x, y)=\frac{x^{3}}{3}-\frac{y^{3}}{3}+9 x y \) What are the critical points? Select the correct choice below and fill in any answer boxes within your choice. A. The critical point(s) is/are (Type an ordered pair. Use a comma to separate answers as needed.) B. There are no critical points.

Ask by Morgan Henry. in the United States
Mar 18,2025

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The critical points are \( (0, 0) \) and \( (9, -9) \).

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Beyond the Answer

To find critical points for the function \( f(x, y)=\frac{x^{3}}{3}-\frac{y^{3}}{3}+9 x y \), we need to find the partial derivatives and set them to zero. 1. Calculate the partial derivative with respect to \( x \): \[ f_x = x^2 + 9y \] 2. Calculate the partial derivative with respect to \( y \): \[ f_y = -y^2 + 9x \] Next, we set both partial derivatives equal to zero: 1. From \( f_x = 0 \): \[ x^2 + 9y = 0 \quad (1) \] 2. From \( f_y = 0 \): \[ -y^2 + 9x = 0 \quad (2) \] From equation (1), we get: \[ 9y = -x^2 \quad \Rightarrow \quad y = -\frac{x^2}{9} \] Substituting \( y \) into equation (2): \[ -\left(-\frac{x^2}{9}\right)^2 + 9x = 0 \] \[ -\frac{x^4}{81} + 9x = 0 \] Multiplying by -81: \[ x^4 - 729x = 0 \] Factoring out \( x \): \[ x(x^3 - 729) = 0 \] Setting each factor to zero gives: 1. \( x = 0 \) 2. \( x^3 = 729 \) \( \Rightarrow x = 9 \) For \( x = 0 \): From (1): \( y = 0 \) For \( x = 9 \): From (1): \[ y = -\frac{9^2}{9} = -9 \] Thus, the critical points are \( (0, 0) \) and \( (9, -9) \). A. The critical point(s) is/are (0, 0), (9, -9)

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