Differentiation rules

Differentiation rules > Using the derivative in modeling

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123456Using the derivative in modeling

Solutions to the exercises

Exercise 1

Just do it.

Your answer.

$A (x) = (x - 2) (\frac{100}{x} - 3)$

$A' (x) = -3 + \frac{200}{x^{2}} = 0$ gives $x^{2} = \frac{200}{3}$ and so $x \approx 8,2$ dm.

The poster should be approximately $8,2$ by $12,2$ dm.

Exercise 2

$Δ A B C$ is uniform with $Δ A D E$ , so $\frac{x}{x + 1} = \frac{3}{D E}$ so that $D E = \frac{3 x + 3}{x} = 3 + \frac{3}{x}$ .
The length of the ladder is $L (x) = \sqrt{{(x + 1)}^{2} + {(3 + \frac{3}{x})}^{2}}$ .
Using differentiation you now determine the minimum of $l (x) = {(x + 1)}^{2} + {(3 + \frac{3}{x})}^{2}$ .
You find a minimal length of $7,56$ m.

Exercise 3

Denote the base of the isosceles triangle by $x$ , then the edges are $10 - \frac{1}{2} x$ each.
The area is $A (x) = \frac{1}{2} x \sqrt{{(10 - \frac{1}{2} x)}^{2} - {(\frac{1}{2} x)}^{2}} = \frac{1}{2} x \sqrt{100 - 10 x}$ .
$A' (x) = \frac{1}{2} \sqrt{100 - 10 x} - \frac{2,5 x}{\sqrt{100 - 10 x}} = 0$ gives $100 - 10 x = 5 x$ and so $x = 6 \frac{2}{3}$ .
So the three edges are all $6 \frac{2}{3}$ cm.

Exercise 4

Make a sketch of the situation.

$A' (k) = \sqrt{10 - 2 k} - \frac{k}{\sqrt{10 - 2 k}} = 0$ gives $\sqrt{10 - 2 k} = \frac{k}{\sqrt{10 - 2 k}}$ and $10 - 2 k = k$ so that $k = 3 \frac{1}{3}$ .

Exercise 5

If $p = 1$ is $f (x) = \frac{x^{2} + 1}{x} = x + \frac{1}{x}$ .
$f' (x) = 1 - \frac{1}{x^{2}} = 0$ gives $x^{2} = 1$ and so $x = -1 \lor x = 1$ . Extrema max. $f (-1) = -2$ and min. $f (1) = 2$ .

$f' (x) = 1 - \frac{p}{x^{2}} = 0$ gives $x^{2} = p$ .
There are no solutions if $p < 0$ and also if $p = 0$ there are no extrema.

$f' (0) = 1 - \frac{p}{x^{2}}$ and $f' (2) = 1 - \frac{p}{4} = -1$ gives $p = 8$ .

Exercise 6

The length of $O P$ is $L (p) = \sqrt{p^{2} + {(4 - p^{2})}^{2}} = \sqrt{p^{4} - 7 p^{2} + 16}$ .
$L (p)$ is minimal if $l (p) = p^{4} - 7 p^{2} + 16$ is minimal.
$l' (p) = 4 p^{3} - 14 p = 0$ if $p = 0 \lor p = \pm \sqrt{3,5}$ .
The minimal length of line segment $O P$ is $L (\pm \sqrt{3,5}) = \sqrt{3,75}$ .

The area of rectangle $A P Q B$ is $A (p) = 2 p (4 - p^{2}) = 8 p - 2 p^{3}$ .
$A' (p) = 8 - 6 p^{2} = 0$ als $p = \pm \sqrt{\frac{4}{3}}$ .
The maximum area is $5 \frac{1}{3} \sqrt{\frac{4}{3}}$ .

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