Page:Calculus Made Easy.pdf/119

Try another simple problem in maxima and minima. Suppose you were asked to divide any number into two parts, such that the product was a maximum? How would you set about it if you did not know the trick of equating to zero? I suppose you could worry it out by the rule of try, try, try again. Let $60$ be the number. You can try cutting it into two parts, and multiplying them together. Thus, $50$ times $10$ is $500$; $52$ times $8$ is $416$; $40$ times $20$ is $800$; $45$ times $15$ is $675$; $30$ times $30$ is $900$. This looks like a maximum: try varying it. $31$ times $29$ is $899$, which is not so good; and $32$ times $28$ is $896$, which is worse. So it seems that the biggest product will be got by dividing into two equal halves.

Now see what the calculus tells you. Let the number to be cut into two parts be called $n$. Then if $x$ is one part, the other will be $n-x$, and the product will be $x(n-x)$ or $nx-x^{2}$. So we write $y=nx-x^{2}$. Now differentiate and equate to zero;

${\dfrac {dy}{dx}}=n-2x=0$.

Solving for $x$, we get ${\dfrac {n}{2}}=x$.

So now we know that whatever number n may be, we must divide it into two equal parts if the product of the parts is to be a maximum; and the value of that maximum product will always be $={\tfrac {1}{4}}n^{2}$.

This is a very useful rule, and applies to any number of factors, so that if $m+n+p=a$ constant number, $m\times n\times p$ is a maximum when $m=n=p$.