Greatest Possible S...
Algebra
Level
4
Let P, Q, R & S be four real numbers such that :
P + Q + R + S = 8
PQ + PR + PS + QR + QS + RS = 12
Find the greatest possible value of S.
6 + 3√2
9 + 2√1
15 - 3√2
2 + √18
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1 solution
P+Q+R+S = 8
PQ+PR+PS+QR+QS+RS = 12
P+Q+R = 8-S
PQ + PR + QR = 12 - S(P+Q+R) = 12-S(8-S) = S^2 - 8S + 12
Obviously,
(P-Q)^2 + (Q-R)^2 + (R-P)^2 >/= 0
2(P^2 + Q^2 + R^2) >/= 2PQ + 2QR + 2RP
[P^2 + Q^2 + R^2 >/= PQ + QR + RP] + 2PQ + 2QR + 2RP
(P+Q+R)^2 >/= 3(PQ + QR + RP)
(8-S)^2 >/= 3(S^2 - 8S + 12)
S^2 - 16S + 64 >/= 3S^2 - 24S + 36
2S^2 - 8S - 28 </= 0
S^2 - 4S - 14 </= 0
Solve for S... S = 2 +/- sq. rt of (18)
So, it will go to...
2 - sq. Rt of (18) </= S </= 2 + sq. Rt of (18)
So, The Maximum value of S = 2 + sq. Rt of (18)