Question textDetermine the position and nature of the stationary point of the function:[math: f(x,y)=2+x2+y2−3xy+x+y]The stationary point is aAnswer 1 Question 7[select: , local minimum, local maximum, median, gradient, point of inflection, saddle point, tangent, root, asymptote]located at [math: (x,y)] where[math: x=] Answer 2 Question 7[input] and [math: y=] Answer 3 Question 7[input]多项填空题
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Question textDetermine the position and nature of the stationary point of the function:[math: f(x,y)=2+x2+y2−3xy+x+y]The stationary point is aAnswer 1 Question 7[select: , local minimum, local maximum, median, gradient, point of inflection, saddle point, tangent, root, asymptote]located at [math: (x,y)] where[math: x=] Answer 2 Question 7[input] and [math: y=] Answer 3 Question 7[input]多项填空题

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BlackTom AI · Accepted Answer

Setting the gradient equal to zero gives the stationary point conditions for f(x,y). Relevant concepts: compute first partial derivatives to find critical points, then use the Hessian (second partials) to classify the stationary point. For a two-variable function, Hessian determinant D = f_xx f_yy − (f_xy)^2; D>0 and f_xx>0 ⇒ local minimum, D>0 and f_xx<0 ⇒ local maximum, D<0 ⇒ saddle point. Step-by-step reasoning and calculations: 1. Compute first partial derivatives: f(x,y) = 2 + x^2 + y^2 − 3xy + x + y f_x = ∂f/∂x = 2x − 3y + 1 f_y = ∂f/∂y = 2y − 3x + 1 2. Solve f_x = 0 and f_y = 0 simultaneously: 2x − 3y + 1 = 0 (1) 2y − 3x + 1 = 0 (2) From (1): 2x = 3y − 1 ⇒ x = (3y − 1)/2. Substitute into (2): 2y − 3[(3y − 1)/2] + 1 = 0 Multiply by 2: 4y − 3(3y − 1) + 2 = 0 4y − 9y + 3 + 2 = 0 ⇒ −5y + 5 = 0 ⇒ y = 1. Then x = (3·1 − 1)/2 = (3 − 1)/2 = 1. So the stationary point is at (x,y) = (1,1). 3. Compute second partial derivatives (Hessian entries): f_xx = ∂^2f/∂x^2 = 2 f_yy = ∂^2f/∂y^2 = 2 f_xy = f_yx = ∂^2f/∂x∂y = −3 Hessian determinant D = f_xx f_yy − (f_xy)^2 = 2·2 − (−3)^2 = 4 − 9 = −5. 4. Classification: D = −5 < 0, so the stationary point is a saddle point (since D<0, the critical point is neither a local min nor max). Why the correct answer is correct: solving the gradient equations yields x = 1 and y = 1; the negative Hessian determinant confirms the nature is a saddle point, ruling out local minimum or maximum. Why other options are incorrect (brief): - Local minimum or maximum: would require D>0 (and appropriate sign of f_xx); here D<0 so impossible. - Point of inflection, gradient, tangent, root, asymptote, median: not applicable as classification of a stationary point in multivariable calculus. Conclusion: The stationary point is a saddle point located at (x,y) = (1,1), confirming the coordinates x = 1 and y = 1 as given.