# Managerial Mathematic - Critical Point

by Yaya
(Malaysia)

Critical Point

•Find:

All critical points of:

f(x,y) = 3x² - 4y² - xy - 2

Subject to the following constraint:

2x + y = 21

### Comments for Managerial Mathematic - Critical Point

 Dec 06, 2012 Critical Point by: Staff Answer Part I ``` Primary Function: f(x,y) = 3x² - 4y² - xy - 2 Subject to the following constraint: 2x + y = 21 Solve the equation representing the constraint for y y = 21 - 2x substitute 21 - 2x (representing the constraint) for y in the primary function f(x) = 3x² - 4y² - xy - 2 f(x) = 3x² - 4(21 - 2x)² - x(21 - 2x) - 2 expand the expression and combine like terms: f(x) = 3x² - 4(4x² - 84x + 441) - x(21 - 2x) - 2 f(x) = 3x² - 16x² + 336x - 1764 - x(21 - 2x) - 2 f(x) = 3x² - 16x² + 336x - 1764 + 2x² - 21x - 2 f(x) = 3x² + 2x² - 16x² + 336x - 21x - 1764 - 2 f(x) = (3x² + 2x² - 16x²) + (336x - 21x) + (- 1764 – 2) f(x) = -11x² + 315x - 1766 f(x) is a parabola which opens downward. ``` ----------------------------------------

 Dec 06, 2012 Critical Point by: Staff ---------------------------------------- Part II ``` the general form of a parabola is: y = ax² + bx + c the quadratic formula is: x = (-b/2a) ± [√(b² - 4ac)/2a] the x coordinate of the vertex of the parabola is: x_vertex = (-b/2a) for your parabola: f(x) = -11x² + 315x - 1766 a = -11 b = 315 x_vertex = -315/[2*(-11)] x_vertex = -315/(-22) x_vertex ≈ 14.3181818181818 x_vertex ≈ 14.3182 substitute the value of x_vertex in the original equation to compute the y value of the vertex f(x) = -11x² + 315x – 1766 x_vertex = 315/22 f(x) = -11*(315/22)² + 315*(315/22) - 1766 f(x) = -99225/44 + 99225/22 - 1766 f(x) = 99225/44 - 1766 f(x) = 21521/44 f(x) ≈ 489.1136363636364 f(x) ≈ 489.114 the coordinates of the vertex in x,y format are: (14.3182, 489.114) The critical point is the vertex (there is only one critical point) Final Answer: Critical Point: (14.3182, 489.114), the maximum value ``` ----------------------------------------------------------------------------

 Dec 06, 2012 Critical Point by: Staff ---------------------------------------------------------------------------- ---------------------------------------- Part III You can also find the critical point by taking the 1st derivative of the function for the parabola, setting the derivative equal to zero, and then solving for x: ``` f(x) = -11x² + 315x – 1766 1st derivative d fx = --- (-11x² + 315x – 1766) dx fx = -22x + 315 set the derivative equal to zero fx = 0 0 = -22x + 315 Solve for x 0 + 22x = -22x + 315 + 22x 22x = -22x + 315 + 22x 22x = 315 + 22x - 22x 22x = 315 + 0 22x = 315 22x / 22 = 315 / 22 x * (22 / 22) = 315 / 22 x * (1) = 315 / 22 x = 315 / 22 x = 315 / 22 x_vertex ≈ 14.3181818181818 x_vertex ≈ 14.3182 Calculating the “y” value is exactly the same as before. Now that you know the x coordinate of the vertex, substitute this value in the original equation to compute y substitute the value of x_vertex in the original equation to compute the y value of the vertex f(x) = -11x² + 315x – 1766 x_vertex = 315/22 f(x) = -11*(315/22)² + 315*(315/22) - 1766 f(x) = -99225/44 + 99225/22 - 1766 f(x) = 99225/44 - 1766 f(x) = 21521/44 f(x) ≈ 489.1136363636364 f(x) ≈ 489.114 the coordinates of the vertex in x,y format are: (14.3182, 489.114) Again, the critical point is the vertex (there is only one critical point) Final Answer: Critical Point: (14.3182, 489.114), the maximum value ``` ----------------------------------------

 Dec 06, 2012 Critical Point by: Staff ----------------------------------------Part IV------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------For your information:The constraint (2x + y = 21) given in the problem statement simplified the problem.The introduction of the constraint simplified the primary function so that it contained only the single variable x, rather than two variables (x, y)           f(x,y) = 3x² - 4y² - xy - 2       to           f(x) = -11x² + 315x – 1766However, it is still possible to calculate critical values when no constraint is given.You can also find the critical point by taking the partial 1st derivative of the function with respect to the variable x, and the partial 1st derivative of the function with respect to y, and then solving the resulting system of equations for x and y.The function f(x,y) = 3x² - 4y² - xy - 2 is a 3-dimensional figure, so it’s a little harder to visualize than a parabola. ----------------------------------------

 Dec 06, 2012 Critical Point by: Staff ----------------------------------------Part V ` f(x,y) = 3x² - 4y² - xy - 2partial 1st derivative of the function with respect to the variable x ∂ fx = --- (3x² - 4y² - xy - 2) ∂x fx = 6x - ypartial 1st derivative of the function with respect to the variable y ∂ fx = --- (3x² - 4y² - xy - 2) ∂y fy = -8y – xSolve the system of two equations for x using substitution 6x - y = 0, -8y - x = 0 -8y - x = 0 -8(6x) - x = 0 -48x - x = 0 -49x = 0 49x = 0 49x / 49 = 0 / 49 x * (49 / 49) = 0 / 49 x * (1) = 0 / 49 x = 0 / 49 x = 0Solve for y by substituting 0 for x 6 * 0 - y = 0 0 - y = 0 - y = 0 y = 0There is only one critical point: (0, 0)To determine the concavity of the critical point, apply the second derivative test. D = AC – B² A = fxx B = fxy * fyx C = fyy D = fxx * fyy – fxy * fyx D = fxx * fyy – f²xy 1. If D > 0 and fxx(x0, (y0) > 0, the point is a relative minimum. 2. If D > 0 and fxx(x0, (y0) < 0, the point is a relative maximum. 3. If D < 0, the point is a saddle point. Computation of D A = fxx = 6 B = fxy = -1 C = fyy = -8 D = AC – B² D = 6*(-8) - (-1)² D = -48 - 1 D = -49 < 0 The critical point (0,0) is a saddle point. ` ----------------------------------------

 Dec 07, 2012 Critical Point by: Staff ---------------------------------------- Part VI Thanks for writing. Staff www.solving-math-problems.com