What is the minimal area under the curve of parabola $y=ax-bx^2$ between $x = 0$ and $x=\frac{a}{b}$ such that it passes through (1,1)?

Question

Let $a > b >0$. The area under a parabola $y=ax−bx^2$ between its two roots, $x=0$ and $x=\frac{a}{b}$, is given by $\frac{a^3}{6b^2}$. Determine the $a$ and $b$ such that the parabola passes through the point $(1,1)$ and this area is minimised.

So far I have calculated the maximum value of this parabola to be at $x = \frac{a}{2b}$ giving $y = \frac{a^2}{4b^2}$. The restriction of $(1,1)$ gives that $1 = a-b$.

I don't know where to go from here so that the area under $y$ is minimised. Any help?

Answer 1

Hint. Since $b=a-1>0$, you have to minimize the function $$f(a)=\frac{a^3}{6b^2}=\frac{a^3}{6(a-1)^2}$$ over $(1,+\infty)$. You may also let $a=b+1$ and minimize the function $$g(b)=\frac{a^3}{6b^2}=\frac{(b+1)^3}{6b^2}$$ over $(0,+\infty)$.