Extremum problems: geometry and the box

Calculus, grade 11 (4 units) Lesson 14 of 20Learning

Extremum problems: geometry and the box

Example: an isosceles triangle

The perimeter of an isosceles triangle is 30 cm.
Its base is $\displaystyle {BC}={2x}$ .
What must $\displaystyle x$ be for the area of the triangle to be greatest?

In the solution below we write the height $\displaystyle AD$ with $\displaystyle x$ by Pythagoras’ theorem, and differentiate the area:

Each leg: $\displaystyle {AB}={\frac{30-2x}{2}}={15-x}$ . The height to the base bisects it, so $\displaystyle {BD}={x}$ .
In triangle $\displaystyle ABD$ : $\displaystyle {h}={\sqrt{(15-x)^2-x^2}}={\sqrt{225-30x}}$ .
The interval: $\displaystyle x>0$ , and a leg is longer than half the base: $\displaystyle 15-x>x$ , that is $\displaystyle x<7.5$ .
The area: $\displaystyle {S(x)}={\frac{2x\cdot h}{2}}={x\sqrt{225-30x}}$ .
$\displaystyle {S'(x)}={\sqrt{225-30x}-\frac{30x}{2\sqrt{225-30x}}}={\frac{225-45x}{\sqrt{225-30x}}}$ .
$\displaystyle {S'(x)}={0}$ when $\displaystyle {x}={5}$ ; before 5 $\displaystyle S'>0$ , after 5 $\displaystyle S'<0$ – a maximum.
The base is $\displaystyle {2x}={10}$ , and each leg $\displaystyle {15-5}={10}$ : the triangle with the greatest area is equilateral, and its area is $\displaystyle {S(5)}={5\sqrt{75}}={25\sqrt{3}}$ cm².

Example: an open box

A box with no lid has a square base of side $\displaystyle x$ cm, and its volume is 32 cm³.
The area of the metal it is made of – its surface area – is to be as small as possible.
In the solution below we write the height with $\displaystyle x$ from the volume, and differentiate the surface area:

The volume: $\displaystyle {x^2\cdot h}={32}$ , so $\displaystyle {h}={\frac{32}{x^2}}$ .
The surface area: the base $\displaystyle x^2$ and four sides of $\displaystyle x\cdot h$ : $\displaystyle {S(x)}={x^2+4x\cdot\frac{32}{x^2}}={x^2+\frac{128}{x}}$ , on the interval $\displaystyle x>0$ .
$\displaystyle {S'(x)}={2x-\frac{128}{x^2}}$ ; $\displaystyle {S'(x)}={0}$ when $\displaystyle {x^3}={64}$ , that is $\displaystyle {x}={4}$ .
Before 4 $\displaystyle S'<0$ , after 4 $\displaystyle S'>0$ – a minimum.
$\displaystyle {x}={4}$ , $\displaystyle {h}={2}$ , and the least surface area: $\displaystyle {S(4)}={16+32}={48}$ cm².

The lateral area of a box is its sides alone, without the bases: here $\displaystyle {4xh}={\frac{128}{x}}$ , and in this box – 32 cm².

Exercise 1

On the graph of $\displaystyle {y}={\sqrt{x}}$ we look for the point nearest to the point $\displaystyle A(2.5,0)$ .
A point on the graph is $\displaystyle (x,\sqrt{x})$ , and its distance from $\displaystyle A$ is $\displaystyle {d(x)}={\sqrt{(x-2.5)^2+(\sqrt{x})^2}}$ .

Expand the brackets under the root, and write $\displaystyle d(x)$ on the board.

\displaystyle f(x)=\sqrt{x}

\displaystyle A=(2.5,0)

Differentiate $\displaystyle d$ .

$\displaystyle d(x)=\sqrt{x^{2}-4x+6.25}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {d'(x)}={0}$ .

Work out the shortest distance. Which point on the graph is it?

$\displaystyle d(x)=\sqrt{x^{2}-4x+6.25}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 2

One vertex of a rectangle is $\displaystyle (0,0)$ , and two of its sides are on the axes.
The vertex opposite the origin is on the graph of $\displaystyle {y}={\frac{4}{x}}$ , and $\displaystyle x>0$ .

The sides of the rectangle are $\displaystyle x$ and $\displaystyle \frac{4}{x}$ . Write the perimeter $\displaystyle P(x)$ on the board.

\displaystyle f(x)=\frac{4}{x}

Differentiate $\displaystyle P$ .

$\displaystyle P(x)=2x+\frac{8}{x}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {P'(x)}={0}$ .

Work out the least perimeter. What shape is the rectangle?

$\displaystyle P(x)=2x+\frac{8}{x}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

A closed box has a square base of side $\displaystyle x$ cm, and its volume is 216 cm³.

Write the height with $\displaystyle x$ , and write the surface area $\displaystyle S(x)$ on the board – two bases and four sides.

Differentiate $\displaystyle S$ .

$\displaystyle S(x)=2x^{2}+\frac{864}{x}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {S'(x)}={0}$ .

Work out the least surface area. What shape is the box?

$\displaystyle S(x)=2x^{2}+\frac{864}{x}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

A closed box has a rectangular base: its sides are $\displaystyle x$ and $\displaystyle 2x$ cm.
The volume of the box is 72 cm³.

The height is $\displaystyle {h}={\frac{36}{x^2}}$ . Write the surface area $\displaystyle S(x)$ on the board.

Differentiate $\displaystyle S$ .

$\displaystyle S(x)=4x^{2}+\frac{216}{x}$

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Solve the equation $\displaystyle {S'(x)}={0}$ .

For the box with the least surface area, work out the lateral area.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: calculus, grade 11 (4 units)

Open as a page

Transformations of the graph of $f$

A horizontal shift: $\displaystyle p$ units to the right ( $\displaystyle p>0$ ) or to the left ( $\displaystyle p<0$ )
$\displaystyle y=f(x-p)$
A vertical shift: $\displaystyle k$ units up ( $\displaystyle k>0$ ) or down ( $\displaystyle k<0$ )
$\displaystyle y=f(x)+k$
A reflection in the $\displaystyle x$ -axis
$\displaystyle y=-f(x)$
A reflection in the $\displaystyle y$ -axis
$\displaystyle y=f(-x)$
A vertical stretch ( $\displaystyle a>1$ ) or squeeze ( $\displaystyle 0<a<1$ )
$\displaystyle y=a\cdot f(x)$
Absolute value: the part below the $\displaystyle x$ -axis is reflected above it
$\displaystyle y=|f(x)|$

The rational function

The horizontal asymptote when the numerator and the denominator have the same degree – $\displaystyle a$ and $\displaystyle b$ the coefficients of the highest power
$\displaystyle y=\frac{a}{b}$
The horizontal asymptote when the numerator’s degree is lower than the denominator’s
$\displaystyle y=0$
An even function – the graph is symmetric about the $\displaystyle y$ -axis
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle f(x)$
An odd function – the graph is symmetric about the origin
$\displaystyle f(-x)$ $\displaystyle =$ $\displaystyle -f(x)$

Derivatives and the tangent

The derivative of a power
$\displaystyle \left(x^{n}\right)'=nx^{n-1}$
The derivative of a product
$\displaystyle \left(u\cdot v\right)'$ $\displaystyle =$ $\displaystyle u'\cdot v+u\cdot v'$
The derivative of a quotient
$\displaystyle \left(\frac{u}{v}\right)'$ $\displaystyle =$ $\displaystyle \frac{u'\cdot v-u\cdot v'}{v^{2}}$
The derivative of $\displaystyle \frac{c}{x}$
$\displaystyle \left(\frac{c}{x}\right)'$ $\displaystyle =$ $\displaystyle -\frac{c}{x^{2}}$
The derivative of a root – $\displaystyle u$ a function of $\displaystyle x$
$\displaystyle \left(\sqrt{u}\right)'$ $\displaystyle =$ $\displaystyle \frac{u'}{2\sqrt{u}}$
The equation of the tangent at the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
The slope of the tangent
$\displaystyle m$ $\displaystyle =$ $\displaystyle f'(x_1)$

Investigating a function with its derivative

The function rises
$\displaystyle f'(x)>0$
The function falls
$\displaystyle f'(x)<0$
A possible extremum point – check the sign of $\displaystyle f'$ on each side of it
$\displaystyle f'(x)=0$

Integrals and areas

The integral of a power
$\displaystyle \int x^{n}\,dx$ $\displaystyle =$ $\displaystyle \frac{x^{n+1}}{n+1}+C$ $\displaystyle n\neq-1$
The integral of $\displaystyle \frac{c}{(ax+b)^{2}}$
$\displaystyle \int\frac{c}{(ax+b)^{2}}\,dx$ $\displaystyle =$ $\displaystyle -\frac{c}{a(ax+b)}+C$
The definite integral – $\displaystyle F$ an antiderivative of $\displaystyle f$
$\displaystyle \int_{a}^{b}f(x)\,dx$ $\displaystyle =$ $\displaystyle F(b)-F(a)$
The area between two graphs, from $\displaystyle {x}={a}$ to $\displaystyle {x}={b}$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \int_{a}^{b}\left(f(x)-g(x)\right)dx$ $\displaystyle f(x)\geq g(x)$