Jacky wants to put a fence around her vegetable garden. She decides to fence it off in the shape of a rectangle and only has $\scriptsize 160\text{ m}$ of fencing so she uses a rock wall as the one border of the garden. Calculate the width and length of the garden that corresponds to the largest possible area that Jacky can fence off.

Solution

Step 1: Read the question carefully, draw a diagram if necessary and decide what variables need to be used.

Let the width be $\scriptsize w$ and the length be $\scriptsize l$.

Step 2: Determine which quantity needs to be maximised or minimised. Write a formula for the quantity to be maximised or minimised in terms of the variables.

We are given information about the area and the modified perimeter of the garden. We know that the area of the garden is given by the formula:

$\scriptsize \text{Area}=l\times w$

Write any other equations relating the independent variables in the formula found above. Use these equations to write the quantity to be maximised or minimised (area in this case) as a function of one variable.

The fencing is only required for three sides and the three sides must add up to $\scriptsize \displaystyle 160\text{ m}$. So we can say the perimeter is:

$\scriptsize w+w+l=160$

Rearrange the formula and make $\scriptsize l$ the subject of the formula.

$\scriptsize l=160-2w$

Substitute the expression for $\scriptsize l$ into the formula for the area of the garden. Notice that this formula now contains only one unknown variable.

$\scriptsize \begin{align*}\text{Area}&=w(160-2w)\\&=160w-2{{w}^{2}}\end{align*}$

Step 3: Differentiate with respect of the variable to find the maximum or minimum value.

We are interested in maximising the area of the garden, so we differentiate to get the following:

$\scriptsize \displaystyle \frac{{dA}}{{dw}}={A}'(w)=160-4w$

Step 4: Calculate the stationary point (point on a graph where a minimum or maximum value is found).

To find the stationary point, we set $\scriptsize {A}'=0$ and solve to find the value(s) of $\scriptsize w$ that maximises the area.

$\scriptsize \begin{align*}{A}'(w)&=160-4w\\0&=160-4w\\4w&=160\\w&=40\end{align*}$

Therefore, the width of the garden is $\scriptsize 40\text{ m}$.

Substitute to solve for the length.

$\scriptsize \begin{align*}l&=160-2w\\&=160-2(40)\\&=80\end{align*}$

Therefore, the length of the garden is $\scriptsize 80\text{ m}$.

We can check that these dimensions give a maximum area by finding the second derivative.

For a maximum value $\scriptsize {A}''(w) \lt 0$:

$\scriptsize \begin{align*}{A}'(w) & =160-4w\\\therefore {A}''(w) & =-4\end{align*}$

$\scriptsize \because {A}'' \lt 0$ The maximum area of the garden is reached when $\scriptsize w=40$.

Therefore, a width of $\scriptsize 40\text{ m}$ and a length of $\scriptsize 80\text{ m}$ gives the maximum area of the garden.

A factory has $\scriptsize x$ employees and makes a profit of $\scriptsize P$ rand per week. The relationship between the profit and number of employees is given by $\scriptsize P\left( x \right)=-2{{x}^{3}}+600x+1~000$.

1. Calculate the number of employees needed for the factory to make a maximum profit.
2. Confirm that the value you found in a) will maximise profit.
3. Calculate the maximum profit.

Solution

1. In this example we are given the equation and variables to work with, which will make things a little easier.
   $\scriptsize P\left( x \right)=-2{{x}^{3}}+600x+1~000$
   We can move straight onto the step where we find the derivative and equate it to zero to solve for the x-value that will maximise profit.
   .
   $\scriptsize \begin{align*}{P}'\left( x \right) & =-6{{x}^{2}}+600\\{P}'\left( x \right) & =0\\-6{{x}^{2}}+600 & =0\\{{x}^{2}} & =100\\x & =\pm 10\\\therefore x & =10\end{align*}$
   .
   Since $\scriptsize x$ represents employees, it cannot be negative and we only accept the positive solution for the number of employees that will maximise profit.
2. To check that $\scriptsize x=10$ maximises profit we must find $\scriptsize {P}''(10)$.
   $\scriptsize \begin{align*}{P}'\left( x \right)&=-6{{x}^{2}}+600\\{P}''(x)&=-12x\\{P}''(10)&=-120\end{align*}$
   Since the second derivative is negative this means that $\scriptsize 10$ employees will maximise profit.
3. To find the maximum profit, substitute the value we found for $\scriptsize x$ back into the original equation, which shows the relationship between profit and employees.
   $\scriptsize \begin{align*}P\left( x \right)&=-2{{x}^{3}}+600x+1~000\\P(10)&=-2{{(10)}^{3}}+600(10)+1000\\&=5000\end{align*}$
   Maximum profit is $\scriptsize \displaystyle \text{R }5000$ when there are $\scriptsize 10$ employees.

The diagram below shows a closed metal cylinder with a capacity of $\scriptsize 550\text{ ml}$. The volume of the cylinder is given by the formula $\scriptsize V=\pi {{r}^{2}}h$ where $\scriptsize r$ is the radius and $\scriptsize h$ is the height.

1. Express the height $\scriptsize h$ of the cylinder in terms of $\scriptsize r$.
2. Show that the surface area of cylinder can be written as $\scriptsize S=2\pi {{r}^{2}}+\displaystyle \frac{{1100}}{r}$.
3. Calculate the radius of the cylinder with a minimum surface area.

Solutions

1. We are given the capacity of the cylinder. When the thickness of the walls of a cylinder are insignificant, volume and capacity is basically the same thing.
   $\scriptsize V=\pi {{r}^{2}}h=550$
   We must use the equation for volume to make the height the subject of the formula.
   $\scriptsize \begin{align*}\pi {{r}^{2}}h&=550\\h&=\displaystyle \frac{{550}}{{\pi {{r}^{2}}}}\end{align*}$
2. For any closed container the easiest way to find the surface area is to use the formula:
   $\scriptsize \text{TSA}=2\times \text{area of the base}+\text{perimeter of the base}\times h$
   Since the base of a cylinder is a circle, its area is $\scriptsize \pi {{r}^{2}}$ and the circumference (perimeter) is $\scriptsize 2\pi r$. Substitute these formulae into the equation for total surface area and replace TSA with S.
   $\scriptsize \begin{align*}S&=2\times \pi {{r}^{2}}+2\pi r\times h\\&=2\pi {{r}^{2}}+2\pi rh\end{align*}$
   Next, we need to rewrite the equation in terms of one variable only. So replace $\scriptsize h$ with the value you found in part a) and simplify to get the required expression.
   $\scriptsize \begin{align*}S&=2\times \pi {{r}^{2}}+2\pi r\times h\\&=2\pi {{r}^{2}}+2\pi rh\\&=2\pi {{r}^{2}}+2\pi r\left( {\displaystyle \frac{{550}}{{\pi {{r}^{2}}}}} \right)\\&=2\pi {{r}^{2}}+\displaystyle \frac{{1\text{ }100}}{r}\end{align*}$
3. Calculate the radius of the cylinder with a minimum surface area.
   To find a minimum value we need to find the derivative. In this case, we need to minimise the surface area of material needed.
   $\scriptsize \begin{align*}S&=2\pi {{r}^{2}}+1\text{ }100{{r}^{{-1}}}\\{S}'&=4\pi r-1\text{ }100{{r}^{{-2}}}\end{align*}$
   Note: we have changed to a negative exponent to get rid of the variable in the denominator before deriving, as you were shown in unit 3.
   .
   For a minimum value $\scriptsize {S}'(r)=0$
   $\scriptsize \begin{align*}4\pi r-1\text{ }100{{r}^{{-2}}}&=0\\4\pi r&=\displaystyle \frac{{1100}}{{{{r}^{2}}}}\\4\pi {{r}^{3}}&=1\text{ }100\\{{r}^{3}}&=\displaystyle \frac{{1\text{ }100}}{{4\pi }}\\r&=\sqrt[3]{{\displaystyle \frac{{275}}{\pi }}}\\&=4.44\end{align*}$
   A radius of $\scriptsize 4.44$ units will produce a cylinder with a minimum surface area, in other words, using the least material.