# 5.2 Solving problems

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## Using differential calculus to solve problems

We have seen that differential calculus can be used to determine the stationary points of functions, in order to sketch their graphs. However, determining stationary points also lends itself to the solution of problems that require some variable to be optimised .

For example, if fuel used by a car is defined by:

$f\left(v\right)=\frac{3}{80}{v}^{2}-6v+245$

where $v$ is the travelling speed, what is the most economical speed (that means the speed that uses the least fuel)?

If we draw the graph of this function we find that the graph has a minimum. The speed at the minimum would then give the most economical speed.

We have seen that the coordinates of the turning point can be calculated by differentiating the function and finding the $x$ -coordinate (speed in the case of the example) for which the derivative is 0.

Differentiating [link] , we get: ${f}^{\text{'}}\left(v\right)=\frac{3}{40}v-6$ If we set ${f}^{\text{'}}\left(v\right)=0$ we can calculate the speed that corresponds to the turning point.

$\begin{array}{ccc}\hfill {f}^{\text{'}}\left(v\right)& =& \frac{3}{40}v-6\hfill \\ \hfill 0& =& \frac{3}{40}v-6\hfill \\ \hfill v& =& \frac{6×40}{3}\hfill \\ & =& 80\hfill \end{array}$

This means that the most economical speed is 80 km $·$ hr ${}^{-1}$ .

The sum of two positive numbers is 10. One of the numbers is multiplied by the square of the other. If each number is greater than 0, find the numbers that make this product a maximum.

1. Let the two numbers be $a$ and $b$ . Then we have:

$a+b=10$

We are required to minimise the product of $a$ and $b$ . Call the product $P$ . Then:

$P=a·b$

We can solve for $b$ from [link] to get:

$b=10-a$

Substitute this into [link] to write $P$ in terms of $a$ only.

$P=a\left(10-a\right)=10a-{a}^{2}$

2. The derivative of [link] is: ${P}^{\text{'}}\left(a\right)=10-2a$

3. Set ${P}^{\text{'}}\left(a\right)=0$ to find the value of $a$ which makes $P$ a maximum.

$\begin{array}{ccc}\hfill {P}^{\text{'}}\left(a\right)& =& 10-2a\hfill \\ \hfill 0& =& 10-2a\hfill \\ \hfill 2a& =& 10\hfill \\ \hfill a& =& \frac{10}{2}\hfill \\ \hfill a& =& 5\hfill \end{array}$

Substitute into [link] to solve for the width.

$\begin{array}{ccc}\hfill b& =& 10-a\hfill \\ & =& 10-5\hfill \\ & =& 5\hfill \end{array}$
4. The product is maximised if $a$ and $b$ are both equal to 5.

Michael wants to start a vegetable garden, which he decides to fence off in the shape of a rectangle from the rest of the garden. Michael only has 160 m of fencing, so he decides to use a wall as one border of the vegetable garden. Calculate the width and length of the garden that corresponds to largest possible area that Michael can fence off.

1. The important pieces of information given are related to the area and modified perimeter of the garden. We know that the area of the garden is: $A=w·l$ We are also told that the fence covers only 3 sides and the three sides should add up to 160 m. This can be written as: $160=w+l+l$

However, we can use [link] to write $w$ in terms of $l$ : $w=160-2l$ Substitute [link] into [link] to get: $A=\left(160-2l\right)l=160l-2{l}^{2}$

2. Since we are interested in maximising the area, we differentiate [link] to get: ${A}^{\text{'}}\left(l\right)=160-4l$

3. To find the stationary point, we set ${A}^{\text{'}}\left(l\right)=0$ and solve for the value of $l$ that maximises the area.

$\begin{array}{ccc}\hfill {A}^{\text{'}}\left(l\right)& =& 160-4l\hfill \\ \hfill 0& =& 160-4l\hfill \\ \hfill \therefore 4l& =& 160\hfill \\ \hfill l& =& \frac{160}{4}\hfill \\ \hfill l& =& 40\mathrm{m}\hfill \end{array}$

Substitute into [link] to solve for the width.

$\begin{array}{ccc}\hfill w& =& 160-2l\hfill \\ & =& 160-2\left(40\right)\hfill \\ & =& 160-80\hfill \\ & =& 80\mathrm{m}\hfill \end{array}$
4. A width of 80 m and a length of 40 m will yield the maximal area fenced off.

## Solving optimisation problems using differential calculus

1. The sum of two positive numbers is 20. One of the numbers is multiplied by the square of the other. Find the numbers that make this product a maximum.
2. A wooden block is made as shown in the diagram. The ends are right-angled triangles having sides $3x$ , $4x$ and $5x$ . The length of the block is $y$ . The total surface area of the block is $3600{\mathrm{cm}}^{2}$ .
1. Show that $y=\frac{300-{x}^{2}}{x}$ .
2. Find the value of $x$ for which the block will have a maximum volume. (Volume = area of base $×$ height.)
3. The diagram shows the plan for a verandah which is to be built on the corner of a cottage. A railing $ABCDE$ is to be constructed around the four edges of the verandah. If $AB=DE=x$ and $BC=CD=y$ , and the length of the railing must be 30 metres, find the values of $x$ and $y$ for which the verandah will have a maximum area.

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Source:  OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
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