Question

Consider the four variables A, B, C and D and a function Z of these variables, $Z = 15A^2 - 3B^4 + C + 0.5D$ . It is given that A, B, C and D must be non-negative integers and that all of the following relationships must hold: i) $2A + B \le 2$ ii) $4A + 2B + C \le 12$ iii) $3A + 4B + D \le 15$

If Z needs to be maximised, then what value must D take?

Accepted Answer

B. To maximize $Z = 15A^2 - 3B^4 + C + 0.5D$, we want $B = 0$ (since $-3B^4$ severely penalizes $Z$). If $B = 0$: From (i) $2A \le 2 \implies A = 0 	ext{ or } 1$. Case 1: $A = 1, B = 0$. From (ii) $4(1) + C \le 12 \implies C \le 8 \implies C_{\max} = 8$. From (iii) $3(1) + D \le 15 \implies D \le 12 \implies D_{\max} = 12$. $Z = 15(1)^2 - 0 + 8 + 0.5(12) = 15 + 8 + 6 = 29$. Case 2: $A = 0, B = 0$. $C \le 12, D \le 15 \implies Z = 0 + 12 + 0.5(15) = 19.5$. Thus, $Z$ is maximized when $A = 1, B = 0, C = 8, D = 12$.

Maximizing Linear Combination Function

Explanation

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