Problem of minimization and maximization

Question

Minimize the expression:

$F(a,b) = ab+ab'+a'b$.

(A) $a'+b'$

(B) $a'+b$

(C) $a+b$

(D) $a+b'$

I have no idea how to solve this problem. Also how to maximize the expression and what is the value of maximization.

Answer 1

I understand that you are dealing with boolean expressions and you are asked to minimize the number of terms (2 instead of 3 for any possible answer (A)...(D)). No maximization...

We are going to prove that the right answer is (C).

We start from the following identity obtained by considering all the elementary events ("$a$ true and $b$ true" OR "$a$ true and $b$ false" OR ... etc.) :

$$ab+ab'+a'b+a'b'=1$$

("$1$" has the meaning "always true")

Therefore, $F(a,b)$ can be written :

$$F(a,b)=ab+ab'+a'b=1-a'b'$$

But, for any boolean expression $B$ $1-B=B'$ i.e., "take the opposite value" (True $\leftrightarrow$ False ; if for example $B=0$ then $B'=1$ and reciprocally). As a consequence :

$$F(a,b)=(a'b')'$$

Using one of the De Morgan's laws (https://en.wikipedia.org/wiki/De_Morgan%27s_laws), one gets:

$$F(a,b)=a+b \ \ \ \ \text{Answer (C)}$$

Remark : It is easy to follow or anticipate the different transformations on the following "Karnaugh diagram" : $$\begin{array}{|c|c|}\hline ab&ab'\\ \hline a'b&a'b'\\ \hline \end{array}$$ where the first line corresponds to $a$ and the second to $a'$, whereas the first column corresponds to $b$ and the second one to $b'$.

Answer 2

Given that this is multiple choice, the fastest approach might be to try some simple values for $a$ and $b$. For example, $F(0,0)=0$, and only choice (C) yields $0$ for $(a,b)=(0,0)$.