Consider point A(6, 30), point B(24, 6) and line AB: 4x+3y = 114.
Point `P(0, lambda)` is a point on y-axis such that `0 lt lambda lt 38 " and point " Q(0, lambda)` is a point on y-axis such that `lambda gt 38`.
For all positions of pont Q, and AQB is maximum when point Q is

Options

A. (0, 54)
B. (0, 58)
C. (0, 60)
D. (0, 1)

Answer

B

Explanation

`"Slope of "AP" is "m_(1)= (30-lambda)/(6)`
`"Slope of "BP" is "m_(2)= (6-lambda)/(24)`
`"tan"theta = ((30-lambda)/(6)-(6-lambda)/(24))/(1+((30-lambda)/(6))((6-lambda)/(24)))`
`=(720-24lambda-36+6lambda)/(144+180-36lambda+lambda^(2))`
`=(18(38-lambda))/(324-36lambda+lambda^(2)) = ((38-lambda)18)/((lambda-18)^(2))`
`"Clearly, for "0 lt lambda lt 38.`
Maximum value of `"tan" theta to oo " for which " theta = (pi)/(2),`
`"Now, " (d)/(dlambda)("tan" theta) = (18(18-lambda)(lambda-58))/((lambda-18)^(4))`
`"Clearly, "lambda = 58` is point of maximum as derivative changes sign from `'+' " to " '-'`.
So, point `Q-=(0,58) " when " angleAQB` is maximum.