Question

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

Answer 1

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`