L with negative slope passes through the point ( 8 , 2 ) and cuts the positive coordinate axes at points P and Q . Find the the absolute minimum value of O P + O Q , as L varies where O is the origin.
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Let m ∈ R + be the slope of line L in question, which is represented as y − 2 = − m ( x − 8 ) ⇒ y = − m x + ( 8 m + 2 ) . The corresponding y and x-intercepts, P and Q , are given as:
P ( 0 , 8 m + 2 ) ; Q ( 8 + m 2 , 0 )
and O P + O Q = L ( m ) = ( 8 m + 2 ) + ( 8 + m 2 ) = 1 0 + 8 m + m 2 . as a function of the slope m . Differentiating L ( m ) produces:
L ′ ( m ) = 8 − m 2 2 = 0 ⇒ 8 m 2 = 2 ⇒ m = 2 1 (since we require m > 0 ) .
and L ′ ′ ( m ) = m 3 4 ⇒ L ′ ′ ( 2 1 ) > 0 (hence, a minimum).
Thus, the minimum value of O P + O Q is L ( 2 1 ) = 1 0 + 8 ( 2 1 ) + 2 1 2 = 1 8 .
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Similar solution with @Dheeman Kuaner 's
Suppose that line L cuts the x - and y -axis at P and Q respectively. Let O P = a , O Q = b and ∠ Q P O = θ . Then we have:
⎩ ⎪ ⎨ ⎪ ⎧ tan θ = a − 8 2 tan θ = a b ⟹ a = tan θ 2 + 8 ⟹ b = a tan θ = 2 + 8 tan θ
Then we have: O P + O Q = a + b = tan θ 2 + 1 0 + 8 tan θ . Using AM-GM inequality tan θ 2 + 8 tan θ ≥ 2 1 6 = 8 . Therefore O P + O Q ≥ 1 8