Calculus or AM-GM?

Algebra Level 3

y = 8 x 3 + 10 x + 1 x 2 y=8x^3+10x+\frac{1}{x^2}

Suppose x > 0 x>0 . Let the minimum of y y be m m . Find the value of ⌊ m ⌋ \lfloor m\rfloor , where ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .

8 7 10 11 9

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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3 solutions

Chan Lye Lee
Oct 27, 2020

We use AM-GM inequality here. Note that y = 8 x 3 + 10 x + 1 x 2 = ( 8 x 3 ) + ( 2 x ) + ( 2 x ) + ( 2 x ) + ( 2 x ) + ( 2 x ) + \displaystyle y=8x^3+10x+\frac{1}{x^2} = \left(8x^3\right)+ \left(2x\right)+ \left(2x\right)+ \left(2x\right)+ \left(2x\right)+ \left(2x\right)+ ( 1 4 x 4 ) + ( 1 4 x 4 ) + ( 1 4 x 4 ) + ( 1 4 x 4 ) ≥ 10 ( ( 8 x 3 ) ( 2 x ) 5 ( 1 4 x 4 ) 4 ) 1 10 = 10 \left(\frac{1}{4x^4}\right)+\left(\frac{1}{4x^4}\right)+\left(\frac{1}{4x^4}\right)+\left(\frac{1}{4x^4}\right) \ge 10 \left( \left(8x^3\right) \left(2x\right)^5\left(\frac{1}{4x^4}\right)^4\right)^{\frac{1}{10}} =10 , with the equality holds if and only if 8 x 3 = 2 x = 1 4 x 4 \displaystyle 8x^3=2x=\frac{1}{4x^4} , that is x = 1 2 \displaystyle x=\frac{1}{2} . Hence the ⌊ m ⌋ = m = 10 \displaystyle \lfloor m\rfloor = m= \boxed{10} .

1 Helpful 1 Interesting 4 Brilliant 0 Confused

Great (if non-obvious?) solution. I had got as far as y = 8 x 3 + 10 x + 1 2 x 2 + 1 2 x 2 ≥ 4 ( 8 x 3 ⋅ 10 x ⋅ 1 2 x 2 ⋅ 1 2 x 2 ) 1 4 = 4 ⋅ 2 0 1 4 ≈ 8.46 y=8x^3+10x+\frac{1}{2x^2}+\frac{1}{2x^2} \ge 4\left(8x^3\cdot 10x \cdot \frac{1}{2x^2} \cdot \frac{1}{2x^2}\right)^{\frac14} = 4\cdot 20^{\frac14}\approx 8.46

...but of course this doesn't quite work. Is there any intuition that can help find solutions like the one you showed?

Chris Lewis - 7 months, 2 weeks ago

@Chris Lewis , while the presentation using AM-GM inequality may look nice, the work behind the scenes may be tedious: Consider y = 8 x 3 + 10 x + 1 x 2 = m ( 8 x 3 m ) + n ( 10 x n ) + p ( 1 p x 2 ) \displaystyle y=8x^3+10x+\frac{1}{x^2} = m\left(\frac{8x^3}{m} \right) + n\left(\frac{10x}{n} \right) +p\left(\frac{1}{px^2} \right) . We need to solve 3 m + n = 2 p \displaystyle 3m+n=2p and 8 x 3 m = 10 x n = 1 p x 2 \displaystyle \frac{8x^3}{m}=\frac{10x}{n} =\frac{1}{px^2} . The intermediate stage we got is 12 x 5 + 5 x 3 = 1 12x^5+5x^3=1 , which is the same if using Calculus. As shown, x = 1 2 x=\frac{1}{2} and one possible answer for ( m , n , p ) (m,n,p) is ( 1 , 5 , 4 ) (1,5,4) .

Chan Lye Lee - 7 months, 2 weeks ago
Chris Lewis
Oct 27, 2020

Here's the calculus version: for a minimum, we need y ′ = 0 y'=0 ; that is 24 x 2 + 10 − 2 x 3 = 0 24x^2+10-\frac{2}{x^3}=0

Rearranging, this is 12 x 5 + 5 x 3 = 1 12x^5+5x^3=1

The unique real root of this is at x = 1 2 x=\frac12 , where we find y = 10 y=\boxed{10} (no need for the floor function). Since the root is unique, this is the only turning point of the function; it's easy to see that y y increases as x → + ∞ x \to +\infty (and as x → 0 + x \to 0^+ ) so the value is indeed the minimum.

2 Helpful 1 Interesting 2 Brilliant 0 Confused
N. Aadhaar Murty
Oct 27, 2020

For minimum, y ′ = 0. y^{\prime} = 0.

∴ 24 x 5 + 10 x 3 − 2 = 0. \therefore 24x^5 +10x^3 -2 = 0.

Graphing this, we can see that x = 0.5 x = 0.5

⇒ y m i n = 10 \Rightarrow y_{min} = 10

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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