Page:Calculus Made Easy.pdf/116

Differentiating, we get:

${\dfrac {dy}{dx}}=2x-4$.

Now equate this to zero, thus:

$2x-4=0$.

Solving this equation for $x$, we get:

$2x$	$=4$,
$x$	$=2$.

Now, we know that the maximum (or minimum) will occur exactly when $x=2$.

Putting the value $x=2$ into the original equation, we get

$y$	$=2^{2}-(4\times 2)+7$
${\displaystyle }$	$=4-8+7$
${\displaystyle }$	$=3$.

Now look back at Fig 26, and you will see that the minimum occurs when $x=2$, and that this minimum of $y=3$.

Try the second example (Fig. 24), which is

$y=3x-x^{2}$.

Differentiating,

${\frac {dy}{dx}}=3-2x$.

Equating to zero,

$3-2x=0$,

whence

$x=1{\tfrac {1}{2}}$;

and putting this value of $x$ into the original equation, we find:

$y$	$=4{\tfrac {1}{2}}-(1{\tfrac {1}{2}}\times 1{\tfrac {1}{2}})$,
$y$	$=2{\tfrac {1}{4}}$.

This gives us exactly the information as to which the method of trying a lot of values left us uncertain.