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AP EAMCET201822 Apr 2018Morning ShiftMathematicsApplication of DerivativesActual

If x and y are two positive numbers such that x+y=32 , then the minimum value of x^2+y^2 is,

Options

  1. A500
  2. B256
  3. C1024
  4. D512

Correct answer

D. 512

Step-by-step solution

aligned & Let, s=x^2+y^2 & =x^2+(32-x)^2 [ x+y=32] & d s d x =2 x+2(32-x)(-1) & =2 x-2(32-x) & =2 x-64+2 x & =4 x-64 & aligned For maxima or minima, array rlrl d s d x & =0 4 x-64 & =0 x & =16 & and y & =32-x=32-16=16 array Again, d^2 s d x =4=+ ive s is minimum when x=y=16 Minimum value of s=16^2+16^2=256+256=512

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