Advertisement Remove all ads

Find the Maxima and Minima of X 3 Y 2 ( 1 − X − Y ) - Applied Mathematics 1

Advertisement Remove all ads
Advertisement Remove all ads
Sum

Find the maxima and minima of `x^3 y^2(1-x-y)`

Advertisement Remove all ads

Solution

We have `f(x)=x63y^2(1-x-y)`

Step 1: `f_x=y^2[3x^2(1-x-y)-x^3]=y^2(3x^2-4x^3-3x^2y)`

`=(3x^2y^2-4x^3y^2-3x^2y^3)`

`f_y=x^3[2y(1-x-y)y(1-x-y)y^2]=x^3(2y-2xy-3y^2)`

`=(2yx^3-2x^4y-3x^3y^2)`

`therefore f_(x x)=6y^2x^1-12x^2y^2-6x^1y^3`

`therefore f_(xy)=6y^1x^2-8x^3y^1-9x^2y^2`

`therefore f_(yy)=2x^3-2x^4-6x^3`

Step 2:- we now solve for `f_y=0 ,f_x=0`

`therefore 3y^2x^2-4x^3y^2-3x^2y^3=0   "i.e."  y^2x^2(3-4x-3y)=0`

And `2y^1x^3-2x^4y^1-3x^3y^2=0   "i.e." y^1x^3(2-2x-3y)=0`

∴ x = 0, y = 0 and (3-4x-3y) = 0, 2-2x-3y = 0

Subtracting we get 1-2x = 0

∴x = ½     ∴3y=3-4(1/2) = 1 `thereforey=1/3`

∴ (0,0) and (1/2,1/3) are stationary points.

Step 3 :- at x=0, y=0, r=0, s=0, t=0 ∴ rt - s2 = 0

At x =½ , y = 1/3

r = f(x x) = 6(1/2)(1/9) - 12(1/4)(1/9) - 6(1/2)(1/27))(1/27)`1/3-1/3-1/0=-1/9`

s = f(xy) = 6(1/4)(1/3) - 8(1/8)(1/3) - 9(1/4)(1/9)9) = 1/2 -`1/3` - 1/4 = - `1/12`

t = f(xy) = 2(1/8) - 2(1/16) - 6(1/8)(1/3) = 1/4 - `1/8` - 1/4 = - `1/8`

∴ rt - s2 = (-1/9)(-1/8)-(1/12)(1/12) = `1/72-1/144=1/144>0`

And r = -1/9 < 0 ∴ f(x,y) is a maxima

Maximum value `=1/8 1/9(1-1/2-1/3)=1/432`

Concept: Maxima and Minima of a Function of Two Independent Variables
  Is there an error in this question or solution?
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×