It is planned to produce X, Y, 180-x-y units respectively in the third quarter.
And should satisfy 40≤x≤ 100, 100≤x+y≤ 180, 0≤y≤ 100, x, y∈N+ (positive integer).
a=50、b=0.2、c=4
Section 1 production cost t 1 = 50x+0.2x 2.
The cost of storing the remaining products into the next quarter is K 1=4(x-40).
Similarly, T2 = 50y+0.2y 2.
K2=4(x+y- 100)
T3 = 50( 180 x-y)+0.2( 180 x-y))^2
Therefore, the total cost f = t1+T2+T3+k1+k2 = 9000+0.2 (x2+y2)+0.2 (180-x-y) 2+4 (2x+y-65433).
manufacture
F'x=0
F'y=0
that is
0.4x-0.4( 180-x-y)+8=0
0.4y-0.4( 180-x-y)+4=0
X=50 y=60。
It is easy to verify at this point.
F''xx≥0
F''yy≥0
Is the minimum point of F.
Compared with the boundary value, it is the smallest point on the domain.
That is, the production plan with the lowest total cost is to produce 50, 60 and 70 units in three quarters respectively.
This is a problem of finding the extreme value in the domain of binary function, so it is more troublesome to do it according to linear programming or nonlinear programming.
As for the influence of A, B and C on the production plan:
The increase or decrease of A has no influence on the production plan at all (no matter how much A is, the plan is 50, 60, 70).
B gradually increasing, the output in three quarters is close to the average of the total delivery, that is, it tends to 60 units (the output in the first quarter increases, the output in the second quarter remains unchanged, and the output in the third quarter decreases).
C gradually increases, and the output in the third quarter is close to the quarterly delivery, that is, it tends to 40, 60 and 80 respectively (the output in the first quarter decreases, the output in the second quarter remains unchanged, and it increases in the third quarter).
God, it's hard to fight. . . . . . . . thank you