Question

The production cost of x units of a product during a month is x ^ 2 dollars. If you want to produce at least 80 units over the next three months, formulate and solve a nonlinear programming model to determine the most economical way to produce.

Answer 1

Consider the given problem here the production cost function is given by, => Ci = Xi^2, for ith month.

The total production of the three month should be at least 80, => X1+X2+X3 > = 80.

So, the cost minimization problem is given by.

=> Minimize “X1^2 + X2^2 + X3^2”, subject to “X1 + X2 + X3 ≥ 80”. So, the Lagrange function is given by.

=> L = {X1^2 + X2^2 + X3^2} + λ*[80 – X1 – X2 – X3]. The first order condition of Kuhn-Tucker conditions are given below.

=> dL/dX1 ≥ 0, X1 ≥ 0, and X1*(dL/dX1) = 0…………..(1).

=> dL/dX2 ≥ 0, X2 ≥ 0, and X2*(dL/dX2) = 0…………..(2).

=> dL/dX3 ≥ 0, X3 ≥ 0, and X3*(dL/dX3) = 0…………..(3).

=> dL/dλ ≤ 0, λ ≥ 0, and λ*(dL/dλ) = 0…………..(4).

Here “X2*(dL/dX2) = 0”, “X2*(dL/dX2) = 0”, “X3*(dL/dX3) = 0” and “λ*(dL/dλ) = 0” are the complementary slackness conditions of the minimization problem.

Let’s assume “X1, X2, X3 and λ > 0”, => to maintain the corresponding complementary slackness conditions the 1^st order partial differential should be zero.

=> dL/dX1 = dL/dX2 = dL/dX3 = dL/dλ = 0.

=> dL/dX1 = 0, => 2*X1 + λ*(-1) = 0, => X1 = λ/2 from condition (1).

=> dL/dX2 = 0, => 2*X2 + λ*(-1) = 0, => X2 = λ/2 from condition (2).

=> dL/dX3 = 0, => 2*X3 + λ*(-1) = 0, => X3 = λ/2 from condition (3).

=> dL/dλ = 0, => X1 + X2 + X3 = 80, from condition (4).

Substituting the values of X1, X2 and X3 into the last equation we have the following expression.

=> X1 + X2 + X3 = 80, => λ/2 + λ/2 + λ/2 = 80, => 3*λ/2 = 80, => 80*2/3 = 160/3, => λ = 160/3.

So, the cost minimizing productions are given by.

=> X1 = λ/2 = (160/3)*(1/2) = (80/3), => X1 = 80/3.

=> X1 = X2 = X3 = 80/3 = 26.67.