Application Of Derivatives

The correct option is(A): 4.

From the information given, we have:

1. ap=1, which implies a=1 (since a and p are positive, and any non-zero number raised to the power of 0 is 1).
2. ab=1, which implies b=a1​=1.
3. pq=1, which means p=q1​ and =p1​.

Now let's focus on the expression to be minimized:

(1+a)(1+b)(1+p)(1+q)

Substitute the values of a, b, p, and q:

(1+1)(1+1)(1+q1​)(1+p1​)

Simplify:

(2)(2)(1+q1​)(1+p1​)

(4)(1+q1​)(1+p1​)

Now, use the information that p and q are positive real numbers and pq=1:

By the AM-GM inequality (Arithmetic Mean-Geometric Mean inequality), for any two positive real numbers x and y, we have:

2x+y​≥xy​

For x=p and y=q, we have:

2p+q​≥pq​

Since pq=1, this simplifies to:

p+q/2​≥1

Now, let's apply this to the expression we want to minimize:

(4)(1+q1​)(1+p1​)=4⋅21+q​⋅21+p​

By the AM-GM inequality:

21+q​≥1⋅q​=q​

21+p​≥1⋅p​=p​

Multiply these inequalities:

1+q​⋅21+p​≥pq​=1

So,

1=44⋅21+q​⋅21+p​≥4⋅1=4

Hence, the least value of (1+a)(1+b)(1+p)(1+q) is 4.