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For $b>a$ what is the maximum possible value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ ?

  • I was trying partial derivative –  May 10 '14 at 13:53
  • And where did that lead you? Can you show your working to indicate where you are stuck? – Calvin Lin May 10 '14 at 13:56
  • Hint: Find the extrema of $f(x)= \dfrac32x-\dfrac{x^2}2-\dfrac{x^3}3$. As your objective is $f(b)-f(a)$, you need to find the difference between a maximum and a minimum to its left. – Macavity May 10 '14 at 14:31

2 Answers2

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By partial differentiation:

Assume $b$ fixed, differentiate with respect to $a$, we get the condition

$$ a^2 + a - \frac {3}{4} = 0 $$

Assume $a$ fixed, differentiate with respect to $b$, we get the condition

$$ b^2 -b - \frac {3}{4} = 0 $$

This gives us $(a, b) = ( - \frac{3}{2}, \frac{1}{2} ) $. It remains to verify if this is indeed a local maximum, and if it is the global maximum. This gives us the value $ \frac{4}{3} $.

Calvin Lin
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For a fixed $a$, the expression is

$$\frac34(b-a)-\frac12(b^2-a^2)-\frac13(b^3-a^3)$$

which is a third degree polynomial.

Differentianting respect to $b$ gives:

$$\frac 34-b-b^2$$ whose roots are $-3/2$ and $1/2$.

The maximum is reached when $b=1/2$. If $a>1/2$, there is no maximum.

ajotatxe
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