A straight line L with negative slope passes through the point (8, 2) and cuts positive coordinate axes at the points P and Q. Find the minimum value of OP + OQ as L varies, where O is the origin.
Answers
Given : A straight line L with negative slope passes through the point (8,2) and cuts the positive coordinate axes at the points P and Q respectively,
To find : minimum value of OP + OQ
Solution:
Let say Line = y = -mx + c m > 0
passes though 8 , 2
=> 2 = -8m + c
=> c = 8m + 2
y = -mx + 8m + 2
P = 0 , 8m + 2
Q = (8m + 2)/ m , 0
OP = 8m + 2
OQ = (8m + 2)/ m
OP + OQ = 8m + 2 + (8m + 2)/ m
L = 8m + 10 + 2/m
dL/dm = 8 - 2/m²
8 - 2/m² == 0
=> m = 1/2 ( m is + ve as -ve sign for slope already taken)
d²L/dm² = 4/m³ > 0
=> L is minimum
Hence OP + OQ is minimum when m = 1/2
OP = 8(1/2) + 2 = 6
OQ = 6/(1/2) = 12
OP + OQ = 6 + 12 = 18 unit
minimum value of OP + OQ 18 units
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