Calculus or AM-GM? (part 2)

Algebra Level 4

U = a + a b U=a+\sqrt{ab}

Suppose that a , b > 0 a,b>0 such that a + b = 100 a+b=100 . Let the maximum value of U U be u u . Find the value of 1000 u \lfloor 1000u\rfloor , where \lfloor \cdot \rfloor denotes the floor function .


The answer is 120710.

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2 solutions

Chan Lye Lee
Nov 1, 2020

As a + b = 100 a+b=100 , we have 1 2 ( 2 + 1 ) ( a + b ) = 1 2 ( 2 + 1 ) 100 \displaystyle \frac{1}{2}\left(\sqrt{2}+1\right) \left(a+b\right) = \frac{1}{2}\left(\sqrt{2}+1\right)100 .

Using AM-GM inequality, 1 2 ( 2 1 ) a + 1 2 ( 2 + 1 ) b 2 ( 1 2 ( 2 1 ) a ) ( 1 2 ( 2 + 1 ) b ) = a b \displaystyle \frac{1}{2}\left(\sqrt{2}-1\right)a + \frac{1}{2}\left(\sqrt{2}+1\right)b \ge 2 \sqrt{ \left(\frac{1}{2}\left(\sqrt{2}-1\right)a\right) \left(\frac{1}{2}\left(\sqrt{2}+1\right)b\right)} = \sqrt{ab} .

Now 1 2 ( 2 + 1 ) 100 = a + 1 2 ( 2 1 ) a + 1 2 ( 2 + 1 ) b a + a b \displaystyle \frac{1}{2}\left(\sqrt{2}+1\right)100 = a + \frac{1}{2}\left(\sqrt{2}-1\right)a + \frac{1}{2}\left(\sqrt{2}+1\right)b \ge a+\sqrt{ab} . The equality holds if and only if 1 2 ( 2 1 ) a = 1 2 ( 2 + 1 ) b \frac{1}{2}\left(\sqrt{2}-1\right)a = \frac{1}{2}\left(\sqrt{2}+1\right)b , which is achievable. Hence u = 1 2 ( 2 + 1 ) 100 u= \frac{1}{2}\left(\sqrt{2}+1\right)100 and thus 1000 u = 120710.678 = 120710 \lfloor 1000u\rfloor = \lfloor 120710.678\rfloor = \boxed{120710} .

Karan Chatrath
Nov 7, 2020

Alternate approach not using AM-GM inequality:

Given that: a + b = 100 a + b = 100 , Let: a = 100 cos 2 x a = 100\cos^2{x} b = 100 sin 2 x b = 100\sin^2{x}

Then the function U U becomes:

U = a + a b U = a + \sqrt{ab} U = 100 cos 2 x + 10000 cos 2 x sin 2 x U = 100\cos^2{x} + \sqrt{10000\cos^2{x}\sin^2{x}} U = 50 ( 1 + cos ( 2 x ) ) + 50 sin ( 2 x ) U = 50(1 + \cos(2x)) + 50\sin(2x) U = 50 + 50 ( cos ( 2 x ) + sin ( 2 x ) ) U = 50 + 50( \cos(2x) + \sin(2x)) U = 50 + 50 2 sin ( 2 x + π 4 ) U = 50 + 50\sqrt{2}\sin \left(2x + \frac{\pi}{4}\right)

So, therefore, U U becomes maximum when: sin ( 2 x + π 4 ) = 1 \sin \left(2x + \frac{\pi}{4}\right)=1 . U m a x = 50 ( 1 + 2 ) \therefore U_{\mathrm{max}} = 50(1 + \sqrt{2})

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