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  • Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

16. Find two positive numbers whose sum is 16 and the sum of whose cubes is
minimum.

Answers (1)

best_answer

let x an d y are positive two numbers 
It is given that 
x + y = 16 , y = 16 - x 
and  x^3 + y^3 is minimum
f(x) = x^3 + (16-x)^3
Now, 
f^{'}(x) = 3x^2 + 3(16-x)^2(-1)
f^{'}(x) = 0\\ 3x^2 - 3(16-x)^2 =0\\ 3x^2-3(256+x^2-32x) = 0\\ 3x^2 -3x^2+96x-768= 0\\ 96x = 768\\ x = 8\\
Hence, x = 8 is the only critical point 
Now,
f^{''}(x) = 6x - 6(16-x)(-1) = 6x + 96 - 6x = 96\\ f^{''}(x) = 96
f^{''}(8) = 96 > 0
Hence, x = 8 is the point of minima
 y = 16 - x
    = 16 - 8 = 8
Hence, values of x and y are 8  and 8 respectively 

Posted by

Gautam harsolia

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