XAT 2020 — QA & DI Question 19
Consider the four variables A, B, C and D and a function Z of these variables, Z = 15A2 − 3B4 + 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 ≤ 2
ii) 4A + 2B + C ≤ 12
iii) 3A + 4B + D ≤ 15
If Z needs to be maximised, then what value must D take?
Answer & solution
- A
15
12
- C
0
- D
10
- E
5
To maximise Z, we need to maximise A, C and D and minimise B.
Given, 2A + B ≤ 2
A can be either 1 or 0.
Case 1: If A = 1, minimum value of B = 0
From (ii): Maximum value of C = 8
From (iii): Maximum value of D = 12
⇒ Z = 15A2 − 3B4 + C + 0.5D = 15 – 0 + 8 + 6 = 29
Case 2: If A = 0, minimum value of B = 0
From (ii): Maximum value of C = 12
From (iii): Maximum value of D = 15
⇒ Z = 15A2 − 3B4 + C + 0.5D = 0 – 0 + 12 + 7.5 = 19.5
We can see that maximum value of Z = 29, when D = 12.
Hence, option (b).