Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to determine the value of a constant $a > 0$ which causes the highest possible value of $f(x) = ax(1-x-a)$.

I have tried deriving the function to find a relation between $x$ and $a$ when $f'(x) = 0$, and found $x = \frac{1-a}{2}$.

I then insert it into the original function: $f(a) = \frac{3a - 6a^2 - 5a^3}{8}$

I derived it to $f'(a) = \frac{-15a^2 + 12a - 3}{8}$

I thought deriving the function and equaling it to $0$ would lead to finding a maximum, but I can't find it. I can't go beyond $-15a^2 + 12a = 3$.

Where am I going wrong?

share|cite|improve this question
One thing that I didn´t understand: $f(a)=\frac {3a-6a^2-5a^3}{8}$ Then,the $f´(a)=\frac {-15x^2+12x-3}{8} $.First is $a$ ,then is $x$? Correct me ,anyone,please,if I´m saying wrong...just did not understand. – HipsterMathematician Jul 20 '12 at 21:17
i think $a$ should be from 2 to infinity – dato datuashvili Jul 20 '12 at 21:25
@MeAndMath Yes, that was a typo. Thanks (and to J.M. who edited). – Quispiam Jul 21 '12 at 8:15
@Quispiam We are here to help! – HipsterMathematician Jul 21 '12 at 20:43
up vote 5 down vote accepted

The first problem is that you substituted $x=\frac12(1-a)$ into $f(x)$ incorrectly; the second (and more important) problem is that you need to define a new function whose independent variable is $a$. Specifically, let $g(a)$ be the maximum attained by the function $f(x)=ax(1-x-a)$; you want to find the value of $a$ that maximizes $g(a)$. Substituting $x=\frac12(1-a)$ into $f(x)$, we find that

$$\begin{align*} g(a)&=a\left(\frac{1-a}2\right)\left(1-\frac{1-a}2-a\right)\\ &=\frac{a}4(1-a)\big(2-(1-a)-2a\big)\\ &=\frac{a}4(1-a)(1-a)\\ &=\frac14\left(a-2a^2+a^3\right)\;. \end{align*}$$

Now $g'(a)=\frac14\left(1-4a+3a^2\right)$. Setting this equal to $0$, we have $3a^2-4a+1=0$. To solve for $a$ you can either use the quadratic formula or notice that $3a^2-4a+1=(3a-1)(a-1)$; either way, you find that $g'(a)=0$ for $a=1$ and $a=\frac13$. By analyzing the sign of $g'(a)$ or by using the second derivative test you can check that $g(a)$ has a local maximum (of $\frac1{27}$) at $a=\frac13$ and a local minimum (of $0$) at $a=1$.

However, a quick check of the graph of $g(a)$ will show you that it increases without bound as $a\to\infty$, and this is also clear algebraically: as $a\to\infty$, $1-a\to-\infty$, so $\frac14a(1-a)^2\to\infty$. Thus, you can make the maximum of $f(x)$ as large as you want by choosing $a$ large enough.

share|cite|improve this answer
@Théophile: Done! – Brian M. Scott Jul 20 '12 at 21:32

The maximum value of the function, which occurs at $x = \frac{1-a}{2}$, is: $$\begin{align}f\big(\frac{1-a}{2}\big) & = a\big(\frac{1-a}{2}\big)(1-\frac{1-a}{2}-a)\\ &= a\big(\frac{1-a}{2}\big)\bigl(\frac{1-a}{2}\bigr)\\ &= a\big(\frac{1-a}{2}\big)^2. \end{align}$$

As $a \rightarrow \infty$, this grows without bound.

share|cite|improve this answer
By the way, note that you don't need calculus to maximize $f(x)$: it is a downward parabola with roots at $0$ and $1−a$, so the $x$-value of the maximum lies halfway between these. – Théophile Jul 20 '12 at 21:29

Before we start differentiating, let's think about $f(x)$ a bit. Let $x=-100$, and let $a=102$. Then $f$ is large. It can be made arbitrarily large by choosing positive $a$ suitably.

If one wants a maximum to exist, the problem needs to be modified. For example, instead of $a\gt 0$, you could specify that $0\lt a\lt 1$, or equivalently that the maximum occurs at a positive value of $x$.

share|cite|improve this answer

Once you insert $x = \dfrac{1-a}{2}$, you can no longer differentiate it since $a$ is a constant and differentiating $f(a)$ will lead to $0$. At best, you can find out $f'(x)$ and substitute the value of x in terms of a to re-verify it is 0 (but then, that's how you got $x = \dfrac{1-a}{2}$ to begin with)

share|cite|improve this answer
No, you can define a new function $g(a)=\max_x f(x)$ and try to maximize $g$. – Brian M. Scott Jul 20 '12 at 21:30

We can resort to some algebra along with the calculus you are using, to see what happens with this function:


Note that this is a parabola. Since the coefficient of the $x^2$ term is negative, it opens downward so that the maximum value is at the vertex. As you have already solved, the vertex has x-coordinate $x=\frac{1−a}{2}$. Additionally, Théophile showed that the vertex's y-coordinate is $f(\frac{1−a}{2})=a(\frac{1−a}{2})^2$ which is unbounded as $a$ increases.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.