Calculus or AM-GM? (part 3)
U = a + a b + 3 a b c + 4 a b c d
Suppose that a , b , c , d > 0 such that a + b + c + d = 1 0 0 . Let the maximum value of U be u . Find the value of ⌊ 1 0 0 0 u ⌋ , where ⌊ ⋅ ⌋ denotes the floor function .
The answer is 142084.
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2 solutions
I am trying to give an idea.Though it is not a complete solution I think the rest part can be easily handled.
we first optimize and make suitable constraints , which is necessary and solve the system of equation (note the 9 variable-equation can be easily simplified using the (iii)constraint to a single variable equation and then solve it. The value of k3''' is coming to be 0.703806842.
Then the answer becomes 142084
I think a sharper bound can be obtained, by some wishful thinking, say the max value is achieved when a = k 2 b = k 4 c = k 6 d , then we have L H S ≤ a + 2 k a + k 2 b + 3 k 2 a + k 2 b + k 4 c + 4 k 3 a + k 2 + k 4 d + k 6 d and we can set the coefficients of each variable on the RHS to be equal, which gives k = 3 2 + 1 3 and this leads to an answer of 163091. Correct me if I'm wrong:).
Very brilliant solution! How did you come up with constraint (iii)?
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To get the maximum value of the desired expression, all three AM/GM inequalities need to be equalities at the same time . Thus we want to be able to have k 1 a = k 2 b = k 3 c = d k 1 ′ a = k 2 ′ b = k 3 ′ c k 1 ′ ′ a = k 2 ′ ′ b and so the desired ratios in (iii) are obtained
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Brilliant! Thanks. This question shows that I did not really understand this AM-Gm inequality.
I don't have a particularly illuminating solution, but here is how I did it.
This is a maximisation subject to a constraint so, we can use Lagrange multiplier and we extremise the following function
U − y ( a + b + c + d − 1 0 0 )
Differentiate w.r.t. a ,b ,c and d gives the following equations to solve.
( 4 y ) 4 = a b c / d 3
( 3 y ) 3 ( c − d ) 3 = a b c
( 2 y ) 2 ( b − c ) 2 = a b
y ( a − b ) = a
And the constraint a + b + c + d = 1 0 0 Before we jump in and solve these nasty beasts, let's note that the maximum U is 100y(just plug in the first 4 eqns.). So, the answer is the floor of 100,000y. Now, to solve for y, I do not have a nice way of doing it. I started by plugging a as a function of y and b and then, find b as a function of c and y and so on. Then I ended up with some messy eqn and I plugged this into Wolfram and it gives y=1.42084 by solving the following eqn
( 3 ( y ( 4 y ( y − 1 ) + 1 ) 2 ) 1 / 3 − 4 1 / 3 ) 3 = 2 7 / ( 6 4 y )
Hello. I would like to learn how did you obtain these four equations? Is y a variable, or a function?
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So, from the last eqn, you can solve a as a function of y and b. Then you plug this into the 3rd eqn, to get b as a function of y and c. Then you plug this into 2nd eqn. Now, you see the pattern.