12.03 Optimisation using calculus

Many practical problems involve finding a solution to achieve a maximum or minimum value such as maximising profit, minimising costs, maximising volume of a box or minimising distance travelled. The process to find the best solution is called optimisation. There are many optimisation techniques for different problems and in many cases where we can describe the outcome with a function we can now use calculus to solve for the optimal solution.

 

Problem solving steps for optimisation applications

1. Where appropriate, draw a clear diagram to illustrate the problem. Label the diagram and define any variables. Note if the variable has any restrictions including implied restrictions - such as the physical dimensions of an object need to be positive.
2. Write an expression for the quantity to be optimised using known relationships. Express this quantity as a function in terms of a single variable - this may require some algebraic manipulation such as substitution.
3. Find the derivative of your function and solve for the value(s) which make the derivative equal zero.
4. Determine the nature of any stationary points found - local maximum, local minimum or stationary point of inflection. (This can be done by checking the value of the derivative either side of any stationary points and it is a good idea to confirm by graphing the function using technology)
5. Identify the optimal solution, remembering to check end points if the domain is restricted and reflect if the solution is practical.

 

Worked examples

example 1

Holly is fencing off an area of her backyard for a vegetable garden. She has $12$12 metres of fencing and is building the garden against an existing wall, so the fencing is only required for three sides. What is the largest area she can enclose with the fencing she has?

Step 1. Draw and label a diagram of the situation.

 

Let $x$x be the width of the garden and $y$y be the length. Since we require the area in terms of one variable, let's rewrite $y$y in terms of $x$x. 

$\text{Total fencing}$Total fencing	$=$=	$2x+y$2x+y
$12$12	$=$=	$2x+y$2x+y
$\therefore y$∴y	$=$=	$12-2x$12−2x

Note: Since both width and length need to be positive we have the implied restriction: $0\le x\le6$0≤x≤6, the end points can be included or not - they will simply give a zero area.

Step 2. We want to maximise area, so we want to express the area in terms of $x$x.

$\text{Area}$Area	$=$=	$\text{length}\times\text{width}$length×width	using known formulas
$A(x)$A(x)	$=$=	$x(12-2x)$x(12−2x)
	$=$=	$12x-2x^2$12x−2x2  $m^2$m2	expanding as this will be required for differentiation

Step 3. Find the derivative and solve $A'(x)=0$A′(x)=0:

The derivative is $A'(x)=12-4x$A′(x)=12−4x, when $A'(x)=0$A′(x)=0, we have:

$12-4x$12−4x	$=$=	$0$0
$4x$4x	$=$=	$12$12
$x$x	$=$=	$3$3

When$x=3$x=3, $A(3)=18$A(3)=18 $m^2$m2. So we have a stationary point at $\left(3,18\right)$(3,18).

Step 4. To confirm this is a maximum we can test the derivative either side of the stationary point:

$x$x	$2$2	$3$3	$4$4
$f'(x)$f′(x)	$4$4	$0$0	$-4$−4
Sign	$+$+	$0$0	$-$−
Shape

Step 5. Hence, we have a maximum area of $18$18 $m^2$m2 when the dimensions of the garden are $3$3 $m$m by $6$6 $m$m.

This could also be confirmed using a graph as seen above or our knowledge of quadratic functions. Recall that graphs of the form $f(x)=ax^2+bx+c$f(x)=ax2+bx+c have their turning point located at $x=-\frac{b}{2a}$x=−b2a​ and this will be a minimum if the leading coefficient $a$a is positive and a maximum if $a$a is negative.

Try using calculus on the general quadratic $f(x)=ax^2+bx+c$f(x)=ax2+bx+c to prove the location of the turning point is always $x=-\frac{b}{2a}$x=−b2a​ and its nature.

Example 2

A piece of wire, $320$320 $cm$cm long, is used to make the twelve edges of a rectangular box in which the length is three times the width. 

a) Find the maximum volume of the box that can be formed using the wire and the dimensions of this optimal box.

Step 1. Draw and label a diagram of the situation.

Let $x$x be the width of the box and $h$h be the height of the box. The length of the box is given by $3x$3x. Since we will require the volume in terms of one variable, let's rewrite $h$h in terms of $x$x. 

$\text{Total length of wire}$Total length of wire	$=$=	$4\left(3x+x+h\right)$4(3x+x+h)
$320$320	$=$=	$4\left(4x+h\right)$4(4x+h)
$4x+h$4x+h	$=$=	$80$80
$\therefore h$∴h	$=$=	$80-4x$80−4x

Note: Since length, width and height need to be positive we have the implied restriction: $0\le x\le20$0≤x≤20

Step 2. We want to maximise the volume, so we want to express the volume in terms of $x$x.

$\text{Volume}$Volume	$=$=	$\text{length}\times\text{width}\times\text{height}$length×width×height	using known formulas
$V(x)$V(x)	$=$=	$3x\times x\times(80-4x)$3x×x×(80−4x)
	$=$=	$240x^2-12x^3$240x2−12x3  $m^2$m2	expanding as this will be required for differentiation

Step 3. Find the derivative and solve $V'(x)=0$V′(x)=0:

The derivative is $V'(x)=480x-36x^2$V′(x)=480x−36x2, when $V'(x)=0$V′(x)=0, we have:

$480x-36x^2$480x−36x2	$=$=	$0$0
$4x\left(120-9x\right)$4x(120−9x)	$=$=	$0$0

Therefore, by null factor law either:

$4x$4x	$=$=	$0$0	or	$120-9x$120−9x	$=$=	$0$0
$\therefore x$∴x	$=$=	$0$0		$-9x$−9x	$=$=	$-120$−120
				$\therefore x$∴x	$=$=	$13\frac{1}{3}$1313​

When$x=0$x=0, $V(0)=0$V(0)=0 $cm^3$cm3 and when $x=13\frac{1}{3}$x=1313​, $V(13\frac{1}{3})=14222\frac{2}{9}$V(1313​)=1422229​ $cm^3$cm3. The volume of $0$0 $m^3$m3 is clearly not a maximum and in fact we do not have a box when the width and length are $0$0 $cm$cm. So let's simply check the nature of the stationary point at $\left(13\frac{1}{3},14222\frac{2}{9}\right)$(1313​,1422229​).

Step 4. To confirm this is a maximum we can test the derivative either side of the stationary point:

$x$x	$10$10	$13\frac{1}{3}$1313​	$15$15
$f'(x)$f′(x)	$1200$1200	$0$0	$-900$−900
Sign	$+$+	$0$0	$-$−
Shape

Step 5. Hence, we have a maximum volume of $14222.\overline{2}$14222.2 $m^3$m3 when the dimensions of the box are $13\frac{1}{3}$1313​ $cm$cm by $40$40 $cm$cm by $26\frac{2}{3}$2623​ $cm$cm. This can also be confirmed by graphing the function on the appropriate domain.

b) If the height of the box can be at most $20$20 $cm$cm, what is the maximum volume that can be formed using the wire and the dimensions of this optimal box.

If the height is restricted we have the following inequality:

$0$0	$\le$≤	$80-4x$80−4x	$\le$≤	$20$20
$-80$−80	$\le$≤	$-4x$−4x	$\le$≤	$-60$−60
$20$20	$\ge$≥	$x$x	$\ge$≥	$15$15
$\therefore15$∴15	$\le$≤	$x$x	$\le$≤	$20$20

Sketching the graph on the restricted domain we can see the maximum occurs at the end point, $x=15$x=15.

Hence, with the height restriction the maximum volume of the box is $13500$13500 $cm^3$cm3, when the dimensions are $15$15 $cm$cm by $45$45 $cm$cm by $20$20 $cm$cm.

 

Practice questions

Question 1

Question 2

Question 3