# Minimum of $\frac{{(b-a)} ^{3}}{6}$

I need help with this problem. Any insight is appreciated. Thanks in advance.

A parabola $y = x^2$ intersects a line at the points $(a,a^2)$ and $(b,b^2)$ $(a < b)$. The area $S$ between the line and the parabola is given by $\frac{{(b-a)} ^{3}}{6}$. Find an expression for the minimum value $m$ of $S$ and the corresponding value of $b$.

• Are there any constraints on $a,b$ other than $a<b$? – πr8 Mar 16 '17 at 0:53
• No, there's just that – unseeingdog Mar 16 '17 at 0:54
• I still do not get it. If $b=a+\varepsilon$, the area between the line and the parabola shrinks to zero, so there is no minimum, and the value of $b$ is irrelevant. – Jack D'Aurizio Mar 16 '17 at 1:02
• The best you can say is that $S$ is bounded below by $0$, i.e. $S > 0$. But this is not the same as minimum value, so we need more information to find a minimum value (a value S can actually obtain). – Cataline Mar 16 '17 at 1:05
• Although there is an expression for $a$ in terms of $b$ – unseeingdog Mar 16 '17 at 1:07

The equation of the line passing through points $(a,a^2)$ and $(b,b^2)$ is $$y = -a b + (a+b)\,x$$ so the area between the line and the curve is $$S=\int_ a^b(-ab+ (a+b)\,x-x^2)\,dx=\frac{(b-a)^3}{6}$$ which is the given result and an identity.
• I was able to find an answer using $a$ = $\frac{(-2b^2-1)}{2b}$, which I got from finding the equation of a line perpendicular to the tangent at any point, which was another part of the problem, but using that would be a limitation for the values of $a$ and $b$, but I am not sure if it also goes in that question. – unseeingdog Mar 16 '17 at 11:20
I think I found the answer. Since $a$ = $\frac{-2b^2-1}{2b}$, $S$ = $\frac{(\frac{4b^2+1}{2b})^3}{6}$. Therefore, $\frac{d}{db}$[$\frac{(\frac{4b^2+1}{2b})^3}{6}$] = 0, and after derivation, $b$ comes out to be either $\frac{1}{2}$ or -$\frac{1}{2}$. Then, using the second derivative test, $b$=-$\frac{1}{2}$ is the the value with negative concavity. Replacing $b$ in $S =$ $\frac{(\frac{4b^2+1}{2b})^3}{6}$, the minimum value turns out to be $-\frac{4}{3}$ and the value of $b$ is $-\frac{1}{2}$. Is that correct?