Maximize $Z = 2X_1 + 5X_2$

Subject to $2X_1 + X_2 ≤ 43 \\ 2X_2 ≤ 46 \\ X_1, X_2 ≥ 0$

Mumbai University > MECH > Sem 7 > Operations Research

Marks: 10 M

Year: May 2014

• Stage I:

  $F_1$ $= \text{Maximize} \ 2X_1 \\ = 2 \max. (X_1)$

  From the constraints:

  The constraint $“2X_2 ≤ 46”$ shows that X1 can go till infinity.

  So we only look at the other constraint.

  $2X_1 + X_2 ≤ 43 → X_1 ≤ \dfrac{43 - X_2}{2} \\ \text{Therefore,} F_1 = 2 \max. \bigg[\dfrac{43 - X_2}{2}\bigg]$

  The maximum value that $X_1$ can assume, is when $X_2$ is zero:

  So $F_1 = 2 × \dfrac{43}{2} = 43$

• Stage II:

  $F_2$ $= \text{Max.} [F_1 + 5X_2] \\ = Max. \bigg[2 \max. \bigg[\dfrac{43 - X_2}{2}\bigg] + 5X_2\bigg]$

  $2X_1 + X_2 ≤ 43 → X_2 ≤ 43 - 2X_1 \\ 2X_2 ≤ 46 →X_2 ≤ 23$

  The max. value of $X_2$ has to be min. $[43 - 2X_1, 23]$

  The maximum value that $X_2$ can assume, is when $X_1$ is zero:

  $Max. X_2 = \min. [43 - 0, 23] = \min. [43, 23] = 23$

  Therefore, $0 ≤ X_2 ≤ 23$ (‘0’ from stage 1, 23 from above)

  Therefore $F_2 = Max. \bigg[2 \bigg\{\dfrac{43 - X_2}{2} \bigg\} + 5X_2\bigg]$

  Since max. $X_2 = 23, \text{putting this value in} \ X_1 ≤ \dfrac{43 - X_2}{2} \ \ give: \ \ X_1= 10$

  Therefore, $F_2 = 2×\dfrac{(43-23)}{2} + 5×23 = 135$

  Therefore, $Z_{\max} = 135, X_1 = 10, X_2 = 23$