Maximum value#3

Algebra Level 5

Let a a and b b be positive real numbers such that a + b ≤ 1 a+b\le 1 . Find the maximum value of

F = a 2 − 3 4 a − a b F={ a }^{ 2 }-\frac { 3 }{ 4a } -\frac { a }{ b }


The answer is -2.25.

Linkin Duck by Linkin Duck , 33, Vietnam

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1 solution

Kushal Bose
Apr 24, 2017

Given a + b ≤ 1 a+b \leq 1 .

Using A.M.-G.M. a + b ≥ 2 a b ⟹ 2 a b ≤ 1 ⟹ a b ≤ 1 / 4 ⟹ b ≤ 1 / 4 a a+b \geq 2 \sqrt{ab} \implies 2 \sqrt{ab} \leq 1 \implies ab \leq 1/4 \implies b \leq 1/4a

b ≤ 1 / 4 a = > 1 b ≥ 4 a = > − 1 b ≤ − 4 a = > − a b ≤ − 4 a 2 b \leq 1/4a => \dfrac{1}{b} \geq 4a => -\dfrac{1}{b} \leq -4a => -\dfrac{a}{b} \leq -4a^2

Using this it can be established F ≤ a 2 − 3 4 a − 4 a 2 = − 3 a 2 − 3 4 a = − ( 3 a 2 + 3 4 a ) = − ( 3 a 2 + 3 8 a + 3 8 a ) ≤ − 3. 3 a 2 . 3 / 8 a . 3 / 8 a 3 = − 9 / 4 = − 2.25 F \leq a^2-\dfrac{3}{4a}-4a^2=-3a^2-\dfrac{3}{4a}=-(3a^2+\dfrac{3}{4a})=-(3a^2+\dfrac{3}{8a}+\dfrac{3}{8a}) \leq - 3.\sqrt[3]{3a^2.3/8a.3/8a}=-9/4=-2.25

So, maximam value of F F is − 2.25 -2.25 which attains when a = b = 1 / 2 a=b=1/2

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