Minimum value

Algebra Level 5

Find the minimum value of the expression a b ( a + b − 28 ) ( a − 1 ) ( b − 27 ) , \frac{ab(a+b-28)}{(a-1)(b-27)}, where a a and b b are positive real numbers such that a > 1 a>1 and b > 27 b>27 .


The answer is 64.

Caleb Garcia Quispe by Caleb Garcia Quispe , 18, Peru

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1 solution

Set A=a-1, b=B-27. Then set B = A c. You will get an expression in A and c:

27 c + 27 A + A ( c + 1 ) + c + 27 c + 28 \frac{\frac{27}{c}+27}{A}+A (c+1)+c+\frac{27}{c}+28

With c fixed, use AM-GM to minimize this expression in A. Now you will have an expression in only c>0 which you have to minimize. After factoring, I got:

( c + 3 3 ) 2 ( c 2 + 1 ) c 2 \frac{\left(c+3 \sqrt{3}\right)^2 \left(c^2+1\right)}{c^2}

I was only able to minimize this function with calculus (at c=Sqrt[3]). You get 64.

Does anybody have a better solution and/or a better way of minimizing this function without calculus?

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Well, you may also add the points where the equality holds.

Aaghaz Mahajan - 3 years, 7 months ago

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