Profit s =10x+9y-c =10x+9y-400-2x-3y-0.01(3x? +xy+3y？ )

The extreme value under the constraint condition x+y= 100, because it is a practical problem, so this extreme value must be the maximum value.

The constructor f(x, y, λ)=S-λ(x+y- 100)

= 10x+9y-400-2x-3y-0.0 1(3x？ +xy+3y？ )-λ(x+y- 100)

F(x, y, λ) takes the partial derivatives of x, y, λ respectively and makes it 0:

-8+0.06 x+0.0 1y-λ= 0( 1)

-6+0.06y+0.0 1x-λ=0 (2)

x+y= 100 (3)

Easy to solve

x=70

y=30