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6]c) Find the maxima and minima of $x^3 y^2 (1-x-y)$
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we have $f(x) = x^3 y^2 (1-x-y)$

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

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

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

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

$∴f_{xx} = 6y^2 x^1-12x^2 y^2-6x^1 y^3$

$∴f_{xy} = 6y^1 x^2-8x^3 y^1-9x^2 y^2$

$∴f_{yy} = 2x^3-2x^4-6x^3$

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

$∴3y^2 x^2-4x^3 y^2-3x^2 y^3 = 0 i.e. y^2 x^2 (3-4x-3y)= 0$

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

$∴ x= 0 , y = 0 and(3-4x-3y)=0 , 2-2x-3y3y 0 $

$Subtracting we get 1-2x = 0$

$∴ x = ½ ∴ 3y = 3-4(1/2) = 1∴y=1/3$

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

$Step 3:- at x = 0, y = 0 , r = 0 , s = 0 , t = 0 ∴ rt - s^2 = 0$

$At x = ½ , y =1/3$

$ r = f_{xx}= 6(1/2)(1/9)-12(1/4)(1/9)-6(1/2)(1/27)/9)-12(1/4)(1/9)-6(1/2)(1/27)1/3- 1/3- 1/9 = - 1/9$

$s = f_{xy} = 6(1/4)(1/3)-8(1/8)(1/3)-9(1/4)(1/9)3)-8(1/8)(1/3)-9(1/4)(1/9) = ½- 1/3- ¼ = -1/12$

$t = f_{xy}= 2(1/8)-2(1/16)-6(1/8)(1/3)(1/16)-6(1/8)(1/3) = ¼- 1/8- ¼ = -1/8$

$∴ rt -s^2 = (-1/9)(-1/8)-(1/12)(1/12)9)(-1/8)-(1/12)(1/12) = 1/72- 1/144 = 1/144\gt0$

And r = -1/9 <0 ∴ f(x,y) is a maxima

Maximum value =1/8.1/9 (1-1/2-1/3) = 1/432

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