We have seen that differential calculus can be effectively used in optimisation problems. That is, if we have a quantity that is varying according to a function (that can be differentiated) we can optimise that quantity by finding maximum and minimum values.
Harder optimisation problems often involve the development of multiple functions which are subsequently reduced down to a single function relating two variables.
The strategy for solving harder optimisation problems is:
- determine the quantity to be maximised or minimised
- determine functions relating the problem variables to each other
- combine the functions to create a single function that relates the quantity to be maximised or minimised to a single variable
- differentiate the function to be optimised and find and classify maximum and minimum points accordingly
- complete a sanity check of your results using the context
As with other optimisation problems, those that involve the trigonometric functions begin with the step of writing functions that model the essential features of the problem. Subsequent steps include finding the derivative and equating it to zero, thus identifying where the gradient is zero.
Because of the periodic nature of the trigonometric functions, there is often an additional complication of choosing which of many possible solutions is or are required.
We also need to remember to undertake a sanity check on the solutions, as they may not make sense for the application in question. For example, a function may give negative values - but if it was representing a volume or an area relationship, those negative values would be nonsensical.
Worked example
Example 1
The graph of the relation $y=\frac{1}{2}x+5$y=12x+5 is drawn between $x=0$x=0 and $x=15$x=15. A rectangle is to be drawn under the graph so that its right-hand edge is at $x=15$x=15, its base is aligned along the $x$x-axis, and its left-hand edge is at a point $x$x.
Find the point $x$x that maximises the area of the rectangle and the value of the area.
Think: We are asked to find the $x$x-coordinate of the point that maximises the area of the rectangle. That is, we need to find a function that relates the area of the rectangle to the variable $x$x.
Do: We will draw a diagram of the situation first and determine enough functions to define our area function $A(x)$A(x).
The area of the rectangle is given by:
$A=bh$A=bh
The length of the base is given by $15-x$15−x and the height is given by the $y$y-coordinate of the $x$x-value that maximises area. Hence the area is given by:
$A=(15-x)y$A=(15−x)y
Currently, we have an equation for area involving two unknowns $x$x and $y$y. However, the function given for the straight line defines a relationship between $x$x and $y$y as follows:
$y=\frac{1}{2}x+5$y=12x+5
Substituting for $y$y in our equation for area, we can deduce that:
$A=\left(\frac{1}{2}x+5\right)(15-x)$A=(12x+5)(15−x)
We now have an equation for area in one unknown $x$x. Note that this equation still accounts for the relationship between $x$x and $y$y. We can differentiate this function using the product rule and then equate to $0$0 to find what possible values of $x$x that maximise $A$A:
| $A'\left(x\right)$A′(x) | $=$= | $\frac{1}{2}\left(15-x\right)+\left(\frac{1}{2}x+5\right)\left(-1\right)$12(15−x)+(12x+5)(−1) |
| $=$= | $\frac{5}{2}-x$52−x |
Equating the gradient function $A'\left(x\right)$A′(x) to $0$0 to find stationary points gives:
| $\frac{5}{2}-x$52−x | $=$= | $0$0 |
| $x$x | $=$= | $\frac{5}{2}$52 |
It is good practice to check the nature of the stationary point to, in this case, confirm it is a maximum. We will do this by determining and evaluating the second derivative at $x=\frac{5}{2}$x=52:
| $A'\left(x\right)$A′(x) | $=$= | $\frac{5}{2}-x$52−x |
| $A''\left(x\right)$A′′(x) | $=$= | $-1$−1 |
In this case the second derivative is always negative, and hence the stationary point at $x=\frac{5}{2}$x=52 must be a maximum turning point.
We can now substitute $x=\frac{5}{2}$x=52 into our area function to find the maximum area:
| $A$A | $=$= | $\left(\frac{1}{2}\cdot\frac{5}{2}+5\right)\left(15-\frac{5}{2}\right)$(12·52+5)(15−52) |
| $=$= | $\frac{625}{8}$6258 | |
| $=$= | $78.125$78.125 |
Practice questions
Question 1
A box without cover is to be constructed from a rectangular cardboard that measures $90$90 cm by $42$42 cm by cutting out four identical square corners of the cardboard and folding up the sides.
Let $x$x be the height of the box, and $V$V the volume of the box.
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Form an equation for $V$V in terms of $x$x.
Give your answer in expanded form.
Solve for the possible value(s) of $x$x that could correspond to the box of largest volume.
Complete the table to prove that when $x=9$x=9 cm the volume of the box is maximised.
$x$x $8$8 $9$9 $10$10 $\frac{dV}{dx}$dVdx $\editable{}$ $0$0 $\editable{}$ Find the maximum volume of the box.
Question 2
In a new mining town, coal needs to be transported from the mine at point $M$M to port at point $P$P. There is an existing train line running $100$100 km from $Q$Q to $P$P such that $M$M is $25$25 km from $Q$Q. A direct road is to be built that connects point $M$M to some part of the train track at point $R$R, a distance of $x$x km from $Q$Q.
Mine operators find that transporting the coal by road costs $\$130$$130 per km, and by rail it costs $\$100$$100 per km.
Find an expression for the total cost of transportation $C$C in terms of $x$x.
Find an equation for the first derivative $\frac{dC}{dx}$dCdx.
Determine the possible value(s) of $x$x that minimise(s) the cost of transporting the coal, where $x>0$x>0. Give your answer correct to the nearest kilometre.
Complete the table to prove that when $x=30$x=30 the cost of transportation in minimised.
Give both values rounded to two decimal places.
$x$x $29$29 $30$30 $31$31 $\frac{dC}{dx}$dCdx $\editable{}$ $0$0 $\editable{}$