Area as a function of a side: the largest area

Science and society, grade 12 (3 units) Lesson 5 of 14Learning

Area as a function of a side: the largest area

A pen with 40 metres of fence

A farm wants to fence a rectangular pen for goats, and has 40 metres of fence.

One side of the rectangle is $\displaystyle x$ metres.
Two neighbouring sides are half the perimeter, 20 metres, so the other side is $\displaystyle 20-x$ metres.
The area, in m²: $\displaystyle {A}={x(20-x)}$ .

$\displaystyle x$ (metres)	$\displaystyle 2$	$\displaystyle 5$	$\displaystyle 8$	$\displaystyle 10$	$\displaystyle 12$	$\displaystyle 15$	$\displaystyle 18$
$\displaystyle A$ (m²)	$\displaystyle 36$	$\displaystyle 75$	$\displaystyle 96$	$\displaystyle 100$	$\displaystyle 96$	$\displaystyle 75$	$\displaystyle 36$

\displaystyle y=x(20-x)

\displaystyle V=(10,100)

With the brackets opened, $\displaystyle {A}={-x^{2}+20x}$ : a quadratic model with $\displaystyle {a}={-1<0}$ , so it has a largest area.
$\displaystyle {x}={-\frac{20}{2\cdot(-1)}}={10}$ , and the other side is $\displaystyle {20-10}={10}$ : the largest pen is a square, 10 by 10, of area 100 m².
A long narrow rectangle, say 2 by 18, takes the same fence and gives only 36 m².

A pen against a wall

If the pen is against the wall of the cowshed, the side along the wall needs no fence.

The two sides at right angles to the wall – $\displaystyle x$ metres each.
The side opposite the wall – $\displaystyle 40-2x$ metres.
The area: $\displaystyle {A}={x(40-2x)}={-2x^{2}+40x}$ .
$\displaystyle {x}={-\frac{40}{2\cdot(-2)}}={10}$ , and the side opposite the wall is $\displaystyle {40-20}={20}$ .

The largest pen is 10 by 20, of area 200 m² – twice the square’s area without the wall. This time it is not a square.

The sides of a given area

Which sides give a pen of 96 m² with 40 metres of fence? We solve the equation:

x(20-x)=96

$\displaystyle {-x^{2}+20x-96}={0}$ , and multiplied by $\displaystyle -1$ : $\displaystyle {x^{2}-20x+96}={0}$ .
$\displaystyle {x}={8}$ or $\displaystyle {x}={12}$ .

Both solutions give the same rectangle: 8 by 12. Once $\displaystyle x$ is the short side, once the long one.

Exercise 1

A school vegetable garden has 60 metres of fence.
One side of the garden is $\displaystyle x$ metres, and its area in m² is $\displaystyle {A}={x(30-x)}$

Mark the graph’s highest point on the board, and check. What are the sides of the largest garden, and what is its area?

\displaystyle y=x(30-x)

Which sides give a garden of 200 m²?

The shorter side

/ – 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

The longer side

/ – 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

A farmer has 24 metres of fence.
He builds a rectangular chicken run against the wall of a shed, and fences only the other three sides.
Each of the two sides at right angles to the wall is $\displaystyle x$ metres.

Write the length of the side opposite the wall, and the run’s area, using $\displaystyle x$ .

Work out the $\displaystyle x$ that gives the largest run.

/ – 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

What is the area of the largest run?

/ – 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

Pupils mark out a rectangular playground with a rope 30 metres long.
Noa says: “The longer the long side, the larger the area”.

Is she right? Explain using the formula of the area or a table.

Formulas: science and society, grade 12 (3 units)

Open as a page

A quadratic model

A quadratic model – $\displaystyle a<0$ : it has a highest point, $\displaystyle a>0$ : a lowest point
$\displaystyle y=ax^{2}+bx+c$ $\displaystyle a\neq0$
The vertex’s $\displaystyle x$ – the axis of symmetry
$\displaystyle x=-\frac{b}{2a}$
The vertex’s $\displaystyle x$ – midway between two values of $\displaystyle x$ with the same $\displaystyle y$
$\displaystyle x=\frac{x_1+x_2}{2}$
The quadratic formula for $\displaystyle {ax^{2}+bx+c}={0}$ – the zeros
$\displaystyle x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

Other models

A linear relation – the differences in the table are equal
$\displaystyle y=mx+b$
An exponential relation – the ratio of two neighbouring values in the table is fixed, $\displaystyle b$
$\displaystyle y=a\cdot b^{x}$

The normal distribution

The standard deviation of $\displaystyle n$ data – $\displaystyle \bar{x}$ their mean
$\displaystyle \sigma=\sqrt{\frac{(x_1-\bar{x})^{2}+(x_2-\bar{x})^{2}+\dots+(x_n-\bar{x})^{2}}{n}}$
In a normal distribution – $\displaystyle \mu$ the mean, $\displaystyle \sigma$ the standard deviation: about 68% of the data within one standard deviation of the mean (34% on each side)
$\displaystyle P(\mu-\sigma<x<\mu+\sigma)\approx68\%$
About 95% within two standard deviations of the mean (another 13.5% on each side)
$\displaystyle P(\mu-2\sigma<x<\mu+2\sigma)\approx95\%$
About 2.5% above $\displaystyle \mu+2\sigma$ , and about 2.5% below $\displaystyle \mu-2\sigma$
$\displaystyle P(x>\mu+2\sigma)\approx2.5\%$
How many of $\displaystyle n$ data are $\displaystyle p$ percent of them
$\displaystyle n\cdot p\%$

The standard score

The standard score of the value $\displaystyle x$ – how many standard deviations it is above the mean (or below it, when $\displaystyle z<0$ )
$\displaystyle z=\frac{x-\mu}{\sigma}$
The value whose standard score is $\displaystyle z$
$\displaystyle x=\mu+z\cdot\sigma$