Question

What is the maximum vertical distance between the line \(y = x + 2\) and the parabola \(y = {x^2}\) for \( - 1 \le x \le 2\)?

Short answer

The maximum distance is \(\frac{9}{4}{\rm{ units}}\).

Step-by-step solution

Optimization of a Function

TheOptimization of a Function\(f\left( x \right)\)for any real value of\(c\)can be given as:

\(\begin{aligned}{l}f\left( x \right)\left| {_{x = c} \to \max \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\\{\rm{and}}\\f\left( x \right)\left| {_{x = c} \to \min \to \left\{ \begin{aligned}{l}{\rm{if }}f'\left( x \right) < 0\,\,\,\,\,\,\forall x < c\\{\rm{if }}f'\left( x \right) > 0\,\,\,\,\,\,\forall x > c\end{aligned} \right.} \right.\end{aligned}\).

Optimizing the function according to given condition.

The given data can be examined with the help of graph as:

Let the vertical distance here be:

\(D\left( x \right) = \left( {x + 2} \right) - {x^2},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ { - 1 \le x \le 2} \right\}\).

For maximization, we have:

\(\begin{aligned}{c}D'\left( x \right) = \frac{d}{{dx}}\left( {\left( {x + 2} \right) - {x^2}} \right)\\\frac{d}{{dx}}\left( {\left( {x + 2} \right) - {x^2}} \right) = 0\\1 - 2x = 0\\x = \frac{1}{2}\end{aligned}\)

Now, we have:

\(\begin{aligned}{c}D\left( { - 1} \right) = 0\\D\left( {\frac{1}{2}} \right) = \frac{9}{4}\\D\left( 2 \right) = 0\end{aligned}\)

Therefore, the function will have absolute maximum value at \(x = \frac{1}{2}\).

Hence, the maximum distance is \(\frac{9}{4}{\rm{ units}}\).