Question

Let ABCD is a square of each side of length 2 cms. M is the midpoint of AB
and ‘P’ is a variable point on BC. Find the smallest possible value of
DP + PM.

Answer 1

Given : ABCD is a square of each side of length 2 cms. M is the midpoint of AB and ‘P’ is a variable point on BC

To find : the smallest possible value of DP + PM.

Solution:

ABCD  is a square of each side of length 2 cms

AB = BC = CD = DA = 2 cm

M is mid point of AB

=> AM = BM = 1 cm

Let say Distance of P from B = x  cm   => BP = x

Then Distance of P from C = 2 - x  cm  => CP = 2 -x

DP² = CD² + CP²  = 2²  + (2 - x)²  = 4 + 4 + x² - 4x    = x² - 4x  + 8

PM²  = BM² + BP²  = 1²  + x²   = x²  + 1

DP + PM =   L  = √ (x² - 4x  + 8)  +  √(x² + 1)

dL/dx  =    (2x - 4)/2√ (x² - 4x  + 8)   +  2x/2√(x² + 1)

=> dL/dx  =    ( x - 2)/√ (x² - 4x  + 8)   +   x/√(x² + 1)

dL/dx  =  0

=>   ( x - 2)/√ (x² - 4x  + 8)   +   x/√(x² + 1)  = 0

=>  ( x - 2)√(x² + 1 ) = - x√ (x² - 4x  + 8)

Squaring on both sides

=> (x² -4x + 4)(x² + 1 ) =  x² (x² - 4x  + 8)

=> x⁴ + x² - 4x³ - 4x + 4x² + 4  = x⁴ - 4x³ + 8x²

=> -3x² - 4x + 4  =0

=>  -3x² - 6x + 2x + 4  =0

=> -3x(x + 2) + 2(x + 2) = 0

=> x = 2/3  or x = -2

Using x  = 2/3  

DP + PM =   3.6  

smallest possible value of  DP + PM.   = 3.6

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