find min value of?

Question

If p and q are non negative real numbers such that p+q=r then the min value of the expression

p+q+2pq
-------------------
1+p+q+pq

(a)

2r
---------
1+r^2

(b)
r
--------
1+r

(c)
2r^2
-------
1+r^2

(d)
2r
----------
1+r

I want to choose the correct option with less time consuming...........can we apply some trick rather than solving the big eqn ?

what would be the answer .

Answer 1

using p + q = r, rewrite the problem as

(r + 2pq)/(1 + r + pq) = (r + pq + pq)/(r + pq + 1)

Now the numerator will be smaller than the denominator when pq < 1. Since both p and q are nonnegative, the smallest pq can be is 0 when either p = 0 or q = 0 or both. When this conditition occurs, the ratio becomes r/(r + 1).

So answer (b).

Math Rules!

Answer 2

The answer is not shown. The correct answer is 1. Unless p and q are DISTINCT, non-negative whole numbers. Otherwise,

(p+q+2pq)/(1+p+q+pq) = (r+2pq)/(r + pq + 1)

If either p or q is greater than 1, 2pq = pq + pq will always be greater than pq + 1 and therefore the ratio will be greater than 1.

Ooop! Non negative, so 0 is a valid value as well. That means the minimum is when both p and q are 0, which would be 0 (not one of the answers. So I'm guessing they have to be distinct, in which case p = 0, q = 1 and the answer would be (b)

Answer 3

Since p + q = r, if we set s = pq we have to minimize f(s) = (r + 2s)/(1 + r + s). If we take the derivative with respect to s, we get

f'(s) = ((1 + r + s)2 - (r + 2s))/(1 + r + s)^2 = (2 + r)/(1 + r + s)^2. Since f'(s) >=0 for every s >0, f is strictly increasing on [0, oo). So, f attains a minimum when s = 0, which implies f(0) = r/(1 + r). Now, since s depends on p and q , we have to chech if s = 0 is a possible value. Since s = pq, we have s = 0 if, and only if, p=0 or q =0, which , in turn, implies p =r and q =0 or p = 0 and q =r. Since these are nonnegative values, s = 0 is possible and the right answer is (b)

Answer 4

p+q+2pq
-------------------
1+p+q+pq
=(r+2pq)/(1+r+pq)
This is at a minimum when pq= 0 so p = 0 or q = 0
So min = r/(1+r)
This assumes that p<> q beause then r = 0 and min value is 0/1 = 0.

Answer 5

We can rewrite the expression as:
[ p(1+q) + q (1+p)]/[(p+1)*(q+1)]
or p/[p+1] + q/[q+1]
Perhaps one of these can help. You haven't stipulated whether p=r.