Average, Mixture and Alligation Question 8 - CAT 2022
If a and b are non-negative real numbers such that a+2b=6, then the average of the maximum and minimum possible values of (a+b)is:
Step-by-Step Explanation
We are given the linear constraint equation a + 2b = 6, where both a and b are constrained to be non-negative real numbers (a ≥ 0, b ≥ 0).
We want to determine the extreme values of the expression a + b.
Express the objective target function entirely in terms of a single variable by substituting a = 6 − 2b:
Target = a + b = 6 − 2b + b = 6 − b
To find the extreme values of 6 − b, find the absolute boundary values for the variable b:
- Since b ≥ 0, the minimum value b can take is 0.
Maximum Value of a + b = 6 − 0 = 6
- Since a ≥ 0, we have 6 − 2b ≥ 0
⟹ 2b ≤ 6 ⟹ b ≤ 3.
Thus, the maximum value b can take is 3.
Minimum Value of a + b = 6 − 3 = 3
The question asks for the average of these maximum and minimum limits:
Average = (6 + 3) / 2 = 4.5
Correct Option: D
💡 FundaMakers Tip:
In linear optimization problem tracks bound by non-negative real constraints, the absolute maximum and minimum limits always occur exactly at the boundary intercepts. You can find them instantly by alternatingly setting each variable to zero. Setting a = 0 gives b = 3 (so a + b = 3), and setting b = 0 gives a = 6 (so a + b = 6). The average of these two boundary conditions gives the final answer immediately.