Question

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

Answer 1

Compute the first partial derivatives

f(x, y) = x³y² (1 − x − y)

(i) Partial derivative w.r.t. x

`f_x = ∂/(∂x) [x^3y^2 (1-x-y)]`

fx ​= 3x²y² (1 − x − y) − x³y²

fx​ = x²y² (3 − 4x − 3y)

(ii) Partial derivative w.r.t. y

`f_y = ∂/(∂y) [x^3y^2 (1-x-y)]`

fy = 2x³y (1 − x − y) − x³y²

fy​ = x³y (2 − 2x − 3y)

fx​ = x²y² (3 − 4x − 3y) = 0

So either

• x = 0, or

• y = 0, or

• 3 − 4x − 3y = 0

fy ​= x³y (2 − 2x − 3y) = 0

3 − 4x − 3y = 0

2 − 2x − 3y = 0

(3 − 4x − 3y) − (2 − 2x − 3y) = 0

1 − 2x = 0 ⇒ x = `1/2`

`2-2 (1/2) - 3y = 0`

2 − 1 − 3y = 0 ⇒ y = `1/3`

`(x, y) = (1/2, 1/3)`

Evaluate the function at this point

`f(1/2, 1/3) = (1/2)^3(1/3)^2 (1-1/2-1/3)`

`f = 1/8 . 1/9 . 1/6`

`=1/864`