# Cubic Equation Word Problems

TR Smith is a product designer and former teacher who uses math in her work every day.

Cubic polynomials are of the form ax^3 + bx^2 + cx + d where a, b, c, and d are coefficients. If the coefficients are real numbers (as opposed to imaginary or complex numbers) a cubic equation ax^3 + bx^2 + cx + d = 0 will either have three real solutions for x, or one real and two complex solutions for x. Finding the roots of a cubic equation is possible by hand, but extremely cumbersome since the formula is not nearly as simple as that of a quadratic equation. In practical applications, you can find the roots with a polynomial solver or graphing calculator, or you can use a convergent numerical algorithm such as Newton's Method. Here are several examples of cubic equations in word problems.

## 1. Thickness of a Fuel Tank's Walls

Measuring from the outside, a metal cylindrical fuel tank has a diameter of 2 feet and the length between its circular ends is 5 feet 6 inches. It is known that the tank can hold 112.25 gallons of fuel. Assuming that the thickness of the metal walls of the tank is uniform, how thick are the tank's walls?

Solution: If the thickness of the tank is uniform, then the space inside the tank is also in the shape of a cylinder. Let x be the thickness of the tank's walls in feet. If the outside diameter of the cylinder is 2 feet and the outside length is 5.5 feet, then the inside diameter of the tank is 2 - 2x and the inside length is 5.5 - 2x. The interior volume in cubic feet is given by the formula

V(x) = (π/4) * (2 - 2x)^2 * (5.5 - 2x)
= π(-2x^3 + 9.5x^2 - 13x + 5.5)

If the capacity is 112.25 gallons, then it holds 15.0056424 cubic feet using the conversion formulas for cubic feet and gallons. Setting the volume equation equal to 15.0056424 gives you

15.0056424 = π(-2x^3 + 9.5x^2 - 13x + 5.5)
2πx^3 - 9.5πx^2 + 13πx - 2.2731172 = 0

Using a cubic equation solver or graphing calculator the approximate roots of this equation are

x ≈ 0.0580943
x ≈ 2.34595 + 0.850842 i
x ≈ 2.34595 - 0.850842 i

The only real-valued solution is the first, x ≈ 0.0581 feet. In inches, this is approximately 11/16 of an inch.

## 2. Cubic Number Puzzle

The product of three distinct numbers is 8400. The sum of the numbers is 72. The difference between the largest and smallest is 32. What are the three numbers?

Solution: Let x be the largest number, y be the mid-sized number, and z be the smallest number. The three equations we have to work with are

xyz = 8400
x + y + z = 72
x - z = 32

The third equation can be rewritten as z = x - 32. If we add the second and third equations together we get 2x + y = 104, which can be rewritten as y = 104 - 2x. Now we can replace z and y with these two expressions to reformulate the first equation.

xyz = x(104 - 2x)(x - 32) = 8400

This is a cubic equation in x. Simplifying gives us

x(104 - 2x)(x - 32) = 8400
-2x^3 + 168x^2 - 3328x = 8400
2x^3 - 168x^2 + 3328x + 8400 = 0
x^3 - 84x^2 + 1664x + 4200 = 0

This cubic equation has three solutions

x = 42
x = 21 + sqrt(541) ≈ 44.2594
x = 21 - sqrt(541) ≈ -2.2594

The last solution must be discarded because if the largest of the three numbers is negative, then all three numbers are negative. But it is impossible for three negative numbers to have a product that is positive. Let's look at the remaining solutions

If x = 42, then using the equations y = 104 - 2x and z = x - 32 we have y = 20 and z = 10. The three numbers 42, 20, and 10 have a product of 8400, a sum of 72,and the difference between the largest and smallest is 32. Therefore, this is one possible solution.

If x = 21 + sqrt(541), then y = 62 - 2*sqrt(541) ≈ 15.4812 and z = -11 + sqrt(541) ≈ 12.2594. This set of three numbers also has a product of 8400, a sum of 72, and the difference between the largest and smallest is also 32. Therefore, this is another possible solution.

Curiously, this problem has two equally valid solution sets. The problem does not say the three numbers must be integers.

## 3. Modeling Physical Processes With Cubic Equations

Jolene is conducting an experiment to see how long it takes for a special kind of paper to burn. She hypothesizes that the time it takes for a piece of this special paper to burn is directly proportional to the area of the piece of paper and inversely proportional to its perimeter. That is, a larger area increases the time to burn, but a larger perimeter decreases the time. After many trials, Jolene has come up with the following equation to predict the time it takes for a piece to burn completely, provided the paper has an area of at least 3 cm^2:

T = 0.23A - 0.08P + 2.49/P

where T is the time in seconds, A is the area in square centimeters, and P is the perimeter in centimeters.

Assuming this formula is correct, how long (to the nearest second) does it take a square piece of paper to burn if the side length is 5 cm? If a square piece of paper takes 16 seconds to burn, about how long are its sides?

Solution: If a square piece of paper is 5 cm along its sides, then it has an area of 25 square cm and a perimeter of 20 cm. This means A = 25 and P = 20. Plugging these into the formula gives us

T = 0.23*25 - 0.08*20 + 2.49/20
= 4.2745 seconds
= 4 seconds to the nearest second.

Now suppose we have a different square piece of paper whose side length is x. The area of this sheet is A = x^2 and the perimeter is P = 4x. If we plug these expressions into the formula we get

T = 0.23x^2 - 0.08(4x) + 2.49/(4x)

If T = 16 seconds, then we get the equation

16 = 0.23x^2 - 0.32x + 0.6225/x

This doesn't look like a cubic equation, but see what happens when we multiply both sides by x:

16x = 0.23x^3 - 0.32x^2 + 0.6225
0.23x^3 - 0.32x^2 - 16x + 0.6225 = 0

Now this a cubic polynomial in standard form. Using a polynomial solver, this cubic has three solutions:

x ≈ 9.0473
x ≈ 0.0389
x ≈ - 7.6949

Only the first solution works in the context of the problem because the formula is only accurate for pieces of at least a certain size and the side length of the paper cannot be negative. Therefore, the piece of paper is about 9 cm along its sides.

## 4. Designing a Lottery With Given Probabilities

Marco is designing a simple random number lottery game in which players select three distinct number from 1 to N. Marco wants to find the value of N so that the probability of a player matching all three numbers is at least 1/200000, but no greater than 1/190000. What should N be?

Solution: If the number set ranges from 1 to N, the probability of matching all three numbers is the reciprocal of the total number of ways to choose three distinct numbers from 1 to N. Therefore, the probability P is

P = 6/[N(N - 1)(N - 2)]

To find the value(s) of N such that 1/200000 ≤ P ≤ 1/190000, we need to solve the equations

1/200000 = 6/[N(N - 1)(N - 2)]
and
1/190000 = 6/[N(N - 1)(N - 2)]

The first equation can be rewritten as

1200000 = N(N - 1)(N - 2)
N^3 - 3N^2 + 2N - 1200000 = 0

This equation has only one positive solution, N ≈ 107.269. The second equation can be rewritten as

1140000 = N(N - 1)(N - 2)
N^3 - 3N^2 + 2N - 1140000 = 0

This has only one positive solution as well, N ≈ 105.468. Analysis of these two equations tells us that N must be an integer greater than 105.468 and less than 107.269. Therefore the possible values for N are 106 and 107.

If Marco's lottery is picking three distinct numbers from 1 to 106, the odds of matching all three are 1 in 192,920. If his lottery is picking three distinct numbers from 1 to 107 the odds of matching all three are 1 in 198,485.

While you could solve this problem with guess-and-check or trial-and-error, such an approach might miss the fact that there are two valid solutions to the problem.

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• SmartAndFun 18 months ago from Texas

Great, Calculus!!! I am dumped with your answering style. I appreciate you.

• Author

TR Smith 18 months ago from Eastern Europe

LOLOLOL! I'm never sure when I'm being mocked or insulted or praised, but I'm glad this hub helped you resolve a cubic equation situation in your life. Stay out of trouble now! For my part, I'm going to try very hard to resist the urge to help people with non-math problems; there's just no helping the witless.

• nicomp really 16 months ago from Ohio, USA

Well done and very interesting. I love simultaneous equations and it's awesome to see them applied to the real world.