Math Problem Statement

giải cho tôi câu này

Given the function f(x, y) = x^2y + 3y^2 - 2xy + 2. Find and classify the function's critical values, separating multiple answers with a comma. If there are no values in a classification, enter DNE.

Solution

To solve the given problem, we need to:

  1. Find the critical points of the function f(x,y)=x2y+3y22xy+2f(x, y) = x^2y + 3y^2 - 2xy + 2.

    • This involves computing the first partial derivatives fxf_x and fyf_y, setting them to zero, and solving for (x,y)(x, y).
  2. Classify the critical points using the second derivative test, which involves:

    • Computing the second partial derivatives fxxf_{xx}, fyyf_{yy}, and fxyf_{xy}.
    • Using the determinant of the Hessian matrix: D=fxxfyy(fxy)2D = f_{xx} f_{yy} - (f_{xy})^2.
      • If D>0D > 0 and fxx>0f_{xx} > 0, the point is a local minimum.
      • If D>0D > 0 and fxx<0f_{xx} < 0, the point is a local maximum.
      • If D<0D < 0, the point is a saddle point.
      • If D=0D = 0, the test is inconclusive.

Let's start by computing the first partial derivatives and finding the critical points. sympy import symbols, diff, solve, hessian, det

Define variables and function

x, y = symbols('x y') f = x2 * y + 3 * y2 - 2 * x * y + 2

Compute first partial derivatives

fx = diff(f, x) fy = diff(f, y)

Solve for critical points (fx = 0 and fy = 0)

critical_points = solve((fx, fy), (x, y)) critical_pointsThe critical points of the function f(x,y)=x2y+3y22xy+2f(x, y) = x^2y + 3y^2 - 2xy + 2 are:

(0,0),(1,16),and(2,0).(0, 0), \quad \left(1, \frac{1}{6}\right), \quad \text{and} \quad (2, 0).

Next, let's classify these points using the second derivative test by calculating the Hessian matrix and its determinant DD. Compute second partial derivatives fxx = diff(fx, x) fyy = diff(fy, y) fxy = diff(fx, y)

Hessian determinant D = fxx * fyy - (fxy)^2

D = fxx * fyy - fxy**2

Evaluate D and fxx at each critical point to classify them

classification = [] for point in critical_points: D_val = D.subs({x: point[0], y: point[1]}) fxx_val = fxx.subs({x: point[0], y: point[1]})

if D_val > 0:
    if fxx_val > 0:
        classification.append((point, 'Local Minimum'))
    elif fxx_val < 0:
        classification.append((point, 'Local Maximum'))
elif D_val < 0:
    classification.append((point, 'Saddle Point'))
else:
    classification.append((point, 'Inconclusive'))

classificationThe classification of the critical points is as follows:

  • (0,0)(0, 0): Saddle Point
  • (1,16)\left(1, \frac{1}{6}\right): Local Minimum
  • (2,0)(2, 0): Saddle Point

Thus, the answers are:

  • Local Maximum: DNE
  • Local Minimum: (1,16)\left(1, \frac{1}{6}\right)
  • Saddle Point: (0,0),(2,0)(0, 0), (2, 0)

Would you like to explore the steps in more detail or have any questions?


Here are 5 related questions to expand your understanding:

  1. How do you compute the Hessian matrix in multivariable calculus?
  2. Why does the sign of the determinant DD classify critical points?
  3. What happens when D=0D = 0 in the second derivative test?
  4. How can a function have no local maxima but still have other critical points?
  5. What are practical applications of finding and classifying critical points?

Tip: Always check your second derivative test calculations carefully to avoid classification errors.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Critical Points
Second Derivative Test

Formulas

f_x = 0 and f_y = 0 to find critical points
D = f_xx f_yy - (f_xy)^2 for Hessian determinant

Theorems

Second Derivative Test for classifying critical points

Suitable Grade Level

Undergraduate