Math urgent :)
posted by Anonymous on .
Hello
A straight line passes through the point (1,27) and intersects the positive xaxis at the point A and the positive yaxis at the point B.Find the shortest possible distance between A and B..
Pleaase could you help me?
Thank you in advance

y = m x + b
when x = 0, y = B so b=B
y = m x + B
when y = 0, x = A
0 = mA+ B
so
m = B/A
so
y = Bx/A + B
when x = 1, y = 27
27 = B/A + B
27 A =  B + BA
27 A = B (A1)
B = 27 A/(A1)
dB/DA = 27 [ (A1)A] /(A1)^2
d^2 = A^2 + B^2
d d^2/dA = 0 for min
= 2 A + 2B dB/DA
so
0 = 2 A + 2B[ 27 [ (A1)A] /(A1)^2
0 = 2A(A1)^2  54 B
0 = 2A(A1)^2 54 [ 27 A/(A1) ]
27 [27/(A1) ] = (A1)^2
27^2 = (A1)^3 = 729
so
A1 = 9
A = 10 amazing. You take it from there. 
oh well curious now
B = 27(10)/9 = 270/9 = 30
d^2 = 10^2 + 30^2 = 1000
d = 10 sqrt(10)
or about 31.6