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2. (14 pts) The two parts below are not related. (a) Find and classify all critical points of \( f(x, y)=3 x y-\frac{1}{2} y^{2}+2 x^{3}+\frac{9}{2} x^{2} \). Clearly show your work and reasoning in using the 2nd derivative test.

Ask by Black Marshall. in the United States
Mar 14,2025

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The critical points of the function \( f(x, y) = 3xy - \frac{1}{2}y^2 + 2x^3 + \frac{9}{2}x^2 \) are \( (0, 0) \) and \( (-3, -9) \). - At \( (0, 0) \), the function has a saddle point. - At \( (-3, -9) \), the function has a local maximum.

Solution

To find and classify the critical points of the function \( f(x, y) = 3xy - \frac{1}{2}y^2 + 2x^3 + \frac{9}{2}x^2 \), we will follow these steps: 1. **Find the first partial derivatives** \( f_x \) and \( f_y \). 2. **Set the first partial derivatives to zero** to find critical points. 3. **Find the second partial derivatives** \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \). 4. **Use the second derivative test** to classify the critical points. ### Step 1: Find the first partial derivatives The first partial derivatives are calculated as follows: \[ f_x = \frac{\partial f}{\partial x} = 3y + 6x^2 + 9x \] \[ f_y = \frac{\partial f}{\partial y} = 3x - y \] ### Step 2: Set the first partial derivatives to zero Now we set the first partial derivatives equal to zero: 1. \( 3y + 6x^2 + 9x = 0 \) (Equation 1) 2. \( 3x - y = 0 \) (Equation 2) From Equation 2, we can express \( y \) in terms of \( x \): \[ y = 3x \] Substituting \( y = 3x \) into Equation 1: \[ 3(3x) + 6x^2 + 9x = 0 \] This simplifies to: \[ 9x + 6x^2 + 9x = 0 \] \[ 6x^2 + 18x = 0 \] Factoring out \( 6x \): \[ 6x(x + 3) = 0 \] This gives us two solutions: 1. \( x = 0 \) 2. \( x = -3 \) Now we can find the corresponding \( y \) values: - For \( x = 0 \): \[ y = 3(0) = 0 \quad \Rightarrow \quad (0, 0) \] - For \( x = -3 \): \[ y = 3(-3) = -9 \quad \Rightarrow \quad (-3, -9) \] Thus, the critical points are \( (0, 0) \) and \( (-3, -9) \). ### Step 3: Find the second partial derivatives Next, we calculate the second partial derivatives: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12x + 9 \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = -1 \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 3 \] ### Step 4: Use the second derivative test The second derivative test involves calculating the determinant \( D \) of the Hessian matrix: \[ D = f_{xx} f_{yy} - (f_{xy})^2 \] Now we evaluate \( D \) at each critical point. #### For the critical point \( (0, 0) \): 1. Calculate \( f_{xx}(0, 0) \): \[ f_{xx}(0, 0) = 12(0) + 9 = 9 \] 2. Calculate \( f_{yy}(0, 0) \): \[ f_{yy}(0, 0) = -1 \] 3. Calculate \( f_{xy}(0, 0) \): \[ f_{xy}(0, 0) = 3 \] 4. Calculate \( D(0, 0) \): \[ D(0, 0) = 9(-1) - (3)^2 = -9 - 9 = -18 \] Since \( D < 0 \), the critical point \( (0, 0) \) is a saddle point. #### For the critical point \( (-3, -9) \): 1. Calculate \( f_{xx}(-3, -9) \): \[ f_{xx}(-3, -9) = 12(-3) + 9 = -36 + 9 = -27 \] 2. Calculate \( f_{yy}(-3, -9) \): \[ f_{yy}(-3, -9) = -1 \] 3. Calculate \( f_{xy}(-3, -9) \): \[ f_{xy}(-3, -9) = 3 \] 4. Calculate \( D(-3, -9) \): \[ D(-3, -9) = (-27)(-1) - (3)^2 = 27 - 9 = 18 \] Since \( D > 0 \) and \( f_{xx}(-3, -9) < 0 \), the critical point \( (-3, -9) \) is a local maximum. ### Summary of Critical Points - The critical point \( (0, 0) \) is a saddle point. - The critical point \( (-3, -9) \) is a local maximum.

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To find the critical points of the function \( f(x, y) = 3xy - \frac{1}{2}y^2 + 2x^3 + \frac{9}{2}x^2 \), we begin by finding the partial derivatives. 1. Calculate the partial derivatives: - \( f_x = \frac{\partial f}{\partial x} = 3y + 6x^2 + 9x \) - \( f_y = \frac{\partial f}{\partial y} = 3x - y \) 2. Set the partial derivatives equal to zero to find critical points: - \( f_x = 0 \) gives us: \( 3y + 6x^2 + 9x = 0 \) \( y = -2x^2 - 3x \) (Equation 1) - \( f_y = 0 \) gives us: \( 3x - y = 0 \) \( y = 3x \) (Equation 2) 3. Substitute Equation 2 into Equation 1: From Equation 1: \( 3x = -2x^2 - 3x \) This simplifies to: \( 2x^2 + 6x = 0 \) Factoring gives: \( 2x(x + 3) = 0 \) Thus, \( x = 0 \) or \( x = -3 \). 4. Find corresponding \( y \) values: - For \( x = 0 \): \( y = 3(0) = 0 \) So one critical point is \( (0, 0) \). - For \( x = -3 \): \( y = 3(-3) = -9 \) So another critical point is \( (-3, -9) \). Now we have the critical points \( (0, 0) \) and \( (-3, -9) \). 5. Classify critical points using the second derivative test: Calculate the second partial derivatives: - \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 12x + 9 \) - \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -1 \) - \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 3 \) Now evaluate at the critical points. **For \( (0, 0) \):** - \( f_{xx}(0, 0) = 12(0) + 9 = 9 \) - \( f_{yy}(0, 0) = -1 \) - \( f_{xy}(0, 0) = 3 \) Calculate the determinant \( D \): \[ D = f_{xx}(0, 0) f_{yy}(0, 0) - (f_{xy}(0, 0))^2 = 9(-1) - 3^2 = -9 - 9 = -18 \] Since \( D < 0 \), \( (0, 0) \) is a saddle point. **For \( (-3, -9) \):** - \( f_{xx}(-3, -9) = 12(-3) + 9 = -36 + 9 = -27 \) - \( f_{yy}(-3, -9) = -1 \) - \( f_{xy}(-3, -9) = 3 \) Calculate the determinant \( D \): \[ D = f_{xx}(-3, -9) f_{yy}(-3, -9) - (f_{xy}(-3, -9))^2 = (-27)(-1) - 3^2 = 27 - 9 = 18 \] Since \( D > 0 \) and \( f_{xx}(-3, -9) < 0 \), \( (-3, -9) \) is a local maximum. In summary, the critical points are: 1. \( (0, 0) \): saddle point 2. \( (-3, -9) \): local maximum

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