Question

p q r are real positive number, such that p+q+r=10 and pq+qr+rp=12 . Then find minimum value of (p+q)??

Answer 1

p+q+r=10 and pq+qr+rp=12 . Then minimum value of p + q = 4/3

Step-by-step explanation:

p+q+r=10

p + q = 10-r

pq+qr+rp=12 .

pq + r(p+q) = 12

pq + r(10-r) = 12

pq = r(r-10) + 12

pq = r^2 - 10r + 12

p + q = 10-r

pq = r^2 - 10r + 12

we can say that p and q are roots of a quadratic equation

x^2 - (10-r)x + r^2 - 10r + 12 = 0

now it's given that

p and q are real

${(10 - r)}^{2} \geqslant 4( {r}^{2} - 10r + 12) \\ 100 + {r}^{2} - 20r \geqslant 4 {r}^{2} - 40r + 48 \\ 3 {r}^{2} - 20r \leqslant 52$

we need to find maximum value of r satisfying this to get minimum value of p + q

as p + q = 10-r

we will get the maximum value of r when

$3 {r}^{2} - 20r \: = 52 \\ 3 {r}^{2} - 20r - 52 = 0$

$3 {r}^{2} + 6r \: - 26r \: - 52 = 0$

3r(r+2) -26(r+2) = 0

(3r-26)(r+2)=0

r = 26/3. (-ve value ignored)

p + q = 10 - 26/3 = 4/3

minimum value of p + q = 4/3

additional info

pq = 4/9

p = 2/3

q = 2/3

r = 26/3