The purpose of this problem is to set the stage for some major applications of Calculus that we will discuss later this semester. The Figure above shows the graph of $$p(x) = x^2(1-x)$$ on the interval $[0,1]$. Clearly the function value assumes a maximum of about 0.15 at a value of $x$ that's close to 0.7.
Within a month or so it will be a very simple matter indeed to figure out the precise maximum value of $p(x)$ and the value of $x$ where that maximum occurs.
If you have taken Calculus before you may already know how to do this, or a tutor can show you how to do this without even breathing hard. However, to get the most benefit from this problem, determine the maximum value of $p(x)$ and the corresponding value of $x$ numerically without the use of Calculus. Use your calculator to evaluate $p$ at many points, or use its graphing facility to zoom in on the maximum. You will appreciate the power of your new skills when you revisit this problem later this semester and solve it using the tools of Calculus. WeBWorK will accept your answer below if it is within one tenth of one percent of the true answer.
$p(x)$ assumes it maximum value of at $x=$ .