Maximizing a in $ax^3+bx^2+cx+d$ with Constraints | POTW #424 07/06/2020

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SUMMARY

The discussion focuses on maximizing the coefficient 'a' in the cubic polynomial function $f(x)=ax^3+bx^2+cx+d$ under the constraint that the absolute value of its derivative, $|f'(x)|$, is less than or equal to 1 for the interval $0 \leq x \leq 1$. The solution provided by user castor28 demonstrates the mathematical approach to derive the maximum value of 'a' while adhering to the specified derivative constraint. This problem exemplifies the application of calculus in optimization scenarios.

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Here is this week's POTW:

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Given that $f(x)=ax^3+bx^2+cx+d$ and $|f'(x)|\le 1$ for $0\le x \le 1$. Find the maximum value of $a$.

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Congratulations to castor28 for his correct solution (Cool) , which you can find below:
We must fit the parabola $y=f'(x)=3x^2+2bx+c$ within the rectangle $[0,1]\times[-1,1]$.

To get the largest value of $a$ (the steepest parabola), we must have $f'(0)=f'(1)=1$, the axis at $x=\dfrac12$, and $f'\left(\dfrac12\right)=-1$:

424s.png


This gives $f'(x)= 8\left(x-\dfrac12\right)^2-1= 8x^2-8x+1$ and $\bf a=\dfrac83$.

More precisely, the condition will be satisfied for $\lvert a\rvert\le\dfrac83$.
 

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