Applications Involving Cubic Functions

Applications of cubic equations in real life are somewhat more scarce than those of quadratic equations. Nevertheless they do occur, particularly in relation to problems involving volume. We discuss three examples here.

Example 1: 

A storage company makes canisters for sporting balls - golf balls, tennis balls basket balls etc. The canisters are cylinders with an opening circular top, and circular base. Irrespective of the ball's size, the canisters are made to hold three balls snugly, so that the diameter of any canister is essentially the diameter of the balls it contains. 

The company is concerned that the packing is inefficient and wants to know the amount of wasted space $W$W measured in $cm^3$cm3, inside a canister of radius $r$r $cm$cm and height $h$h $cm$cm. They also want to develop $W$W as a function of the canisters radius, and determine the waste for tennis balls of radius $r=3.43$r=3.43 $cm$cm and golf balls of radius $r=2.14$r=2.14 $cm$cm.

The volume of a cylinder is given by $V_c=\pi r^2h$Vc​=πr2h and the volume of a sphere is given by $V_s=\frac{4}{3}\pi r^3$Vs​=43​πr3. Therefore $W=\pi r^2h-3\times\left(\frac{4}{3}\pi r^3\right)$W=πr2h−3×(43​πr3). 

The height $h$h of the canister is $6$6 radii, and so substituting $h=6r$h=6r, and simplifying we have:

$W$W	$=$=	$\pi r^2h-3\times\left(\frac{4}{3}\pi r^3\right)$πr2h−3×(43​πr3)
	$=$=	$\pi r^2\times\left(6r\right)-3\times\left(\frac{4}{3}\pi r^3\right)$πr2×(6r)−3×(43​πr3)
	$=$=	$6\pi r^3-4\pi r^3$6πr3−4πr3
	$=$=	$2\pi r^3$2πr3

Note that the wasted space, as a proportion of any canister, is the constant $\frac{2\pi r^3}{6\pi r^3}=\frac{1}{3}$2πr36πr3​=13​.

In particular, the wasted space of a tennis ball container is given by $W_t=2\pi\left(3.43\right)^3=253.55$Wt​=2π(3.43)3=253.55 $cm^3$cm3, and the wasted space of a golf ball container is given by $W_g=2\pi\left(2.14\right)^3=61.58$Wg​=2π(2.14)3=61.58 $cm^3$cm3.

Example 2:

An open chip box is made from a sheet of cardboard measuring $30$30 $cm$cm by $20$20 $cm$cm with four squares cut away from each corner as depicted in the diagram.

Here is the constructed box:

The store owner wishes to know the volume of the box as a function of the side $x$x of the cut-away squares. She suspects that the quantity of chips that can be contained varies according to the size of $x$x, but is not sure.

To answer this question, the length of the box, when constructed, will be $l=30-2x$l=30−2x and the width will likewise be $w=20-2x$w=20−2x.  With the height of the box as $x$x, the volume is given by $V=\left(30-2x\right)\left(20-2x\right)x$V=(30−2x)(20−2x)x or when expanded $V=4x^3-100x^2+600x$V=4x3−100x2+600x. 

Of course we only want positive volumes for our function, so we need to solve the inequality $\left(30-2x\right)\left(20-2x\right)x\ge0$(30−2x)(20−2x)x≥0 to find the restrictions on the domain. 

The inequality, when the common factor $4$4 is taken out becomes $\left(15-x\right)\left(10-x\right)x\ge0$(15−x)(10−x)x≥0 and the critical points are $0$0, $10$10 and $15$15. Testing the four intervals shows the interval we are looking for is given by $0\le x\le10$0≤x≤10. 

The graph of $V$V as a function of $x$x is shown here. 

You can see from the graph that the maximum volume occurs when $x=4$x=4, and this volume is determined as $V=\left(30-2\times4\right)\left(20-2\times4\right)\times4=1056$V=(30−2×4)(20−2×4)×4=1056 $cm^3$cm3. 

Example 3:

John owns an insurance firm. He employs staff to sell insurance policies to people. He finds that if he employs $m$m staff (at a cost of $\$750$$750 per week per staff member) then he sells $q$q policies where:

$q=m^3-12m^2+60m$q=m3−12m2+60m 

Because of space considerations, he cannot hire anymore than $7$7 staff.  He also is up for $\$2500$$2500 a week in fixed costs.

If the firm receives  $\$50$$50 for every policy sold, how many staff should he hire?

To answer this we need to look at maximising the profit, which is the weekly revenue less the weekly costs.

From the information contained in the question we can form a profit function $P$P given as:

$P=50\times\left(m^3-12m^2+60m\right)-750m-2500$P=50×(m3−12m2+60m)−750m−2500

This simplifies to $P=m^3-12m^2-690m-2500$P=m3−12m2−690m−2500, and by using factoring techniques we can express this as $P=50\left(m-5\right)^2\left(m-2\right)$P=50(m−5)2(m−2).

Note that the function's domain (eliminating $0$0 staff) is given by $\left\{1,2,3,4,5,6,7\right\}${1,2,3,4,5,6,7}, and therefore we can evaluate the function using a profit table shown here:

$m$m	$1$1	$2$2	$3$3	$4$4	$5$5	$6$6	$7$7
$P$P	$-800$−800	$0$0	$200$200	$100$100	$0$0	$200$200	$1000$1000

Thus, John needs to employ $7$7 staff for a profit of $\$1000$$1000 a week.

 

More Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3