logo for solving-math-problems.com
leftimage for solving-math-problems.com

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

Click here to add your own comments

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.


Parabola – graph of  f(x) = -11x² + 315x - 1766






----------------------------------------

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 – 1766


However, 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 - 2

partial 1st derivative of the function with respect to the variable x


fx = --- (3x² - 4y² - xy - 2)
∂x



fx = 6x - y

partial 1st derivative of the function with respect to the variable y


fx = --- (3x² - 4y² - xy - 2)
∂y

fy = -8y – x

Solve 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 = 0

Solve for y by substituting 0 for x

6 * 0 - y = 0

0 - y = 0

- y = 0

y = 0


There 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


Math – graph of  f(x,y) = 3x² - 4y² - xy - 2, view 1





Math – graph of  f(x,y) = 3x² - 4y² - xy - 2, view 2








Thanks for writing.

Staff
www.solving-math-problems.com



Click here to add your own comments

Join in and write your own page! It's easy to do. How? Simply click here to return to Math Questions & Comments - 01.



Copyright © 2008-2015. All rights reserved. Solving-Math-Problems.com