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Quadratic Equation Word Problems

Updated on September 06, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

In high school algebra classes, quadratic equations problems are usually presented in the form "Solve x^2 + 3x + 4 = 0" without much context or indication of how this could ever be useful in real life. However, quadratic equations arise naturally in many problems in physics, chemistry, economics, business, statistics, engineering, biology, and architecture. Here are some high school-level and college algebra-level math word problems whose solutions involve quadratic equations. See also the 9-Question Quadratic Equations Quiz for more practice.

(1) Square Tiles -- Challenging Quadratic Equation Problem

Juliet has 96 square tiles. She wants to arrange them in the form of an open rectangle with a dividing line of tiles down the middle that divides the larger rectangle into two smaller rectangular spaces of equal area, as shown in the image below. Moreover, she wants to arrange the tiles so that the combined area of the two spaces is as large as possible. How can she do this without cutting any tiles?

Solution: In the example diagram above, a Juliet-style arrangement is made with 122 tiles. We need to find a solution to the problem using 96 tiles. Let's call the length of the larger rectangle L and let's call the width minus the two tiles at the ends W. Since we are restricted to 96 tiles, we have

2L + 3W = 96

The total area of the two smaller rectangular spaces is given by the equation

Area = (L - 3)W

If we solve the equation 2L + 3W = 96 for L, we get L = 48 - 1.5W. This means we can write the area equation in terms of a single variable, W, as

Area = [(48 - 1.5W) - 3]W
= (45 - 1.5W)W
= 45W = 1.5W^2

As we can see, the area equation is a quadratic function of W. Since the coefficient of the W^2 term is negative, the graph is a parabola opening downward with a maximum value at the vertex. Graphically, the height of the parabola at a particular point W gives the area for that value of W, and the vertex of the parabola is the maximum possible area.

The vertex of a parabola is always halfway between the roots, so what we need to do for this problem is find the roots of 45W - 1.5W^2 and take their average. Solving gives us

45W - 1.5W^2 = 0
W = 0 and W = 30

The halfway point between 0 and 30 is W = 15. However, there is one problem with this solution. If we have W = 15 then we have L = 25.5 -- a fractional value requiring cut tiles, but the constraints of our problem say that Juilet cannot cut the tiles.

If we go on either side of the vertex at W = 14 and W = 16, we get L = 27 and L = 24 respectively. In the first case, the total area of the two rectangular spaces is

Area = (L - 3)W
= (27 - 3)*14
= 24*14
= 336

In the second case the total area of the two rectangular spaces is

Area = (L - 3)W
= (24 - 3)*16
= 21*16
= 336

So it turns out these two cases both give the optimal area. This means Juliet can arrange her square tiles in either of the two arrangements below to get the maximum area.

(2) River Current Speed -- Medium Difficulty

A small motorboat makes a round trip on a river from Bob's riverfront property to Amanda's and back again to Bob's. The two locations are 28 miles apart. Going upstream from Bob's to Amanda's the boat runs at its maximum speed setting of 16 mph relative to still water. Going downstream from Amanda's back to Bob's the boat runs at its lowest setting of 5 mph relative to still water. If the entire trip takes 6 hours, what is the speed of the river's current?

Solution: Call the speed of the river current R. The speed of the boat relative to the river bank going from Bob's to Amanda's is 16 - R because the boat is running against the current. The time it takes to go from Bob's to Amanda's is 28/(16 - R).

The speed of the boat going from Amanda's back to Bob's is 5 + R because the boat is now running with the current. The time it takes to complete this leg of the trip is 28/(5 + R).

If the whole trip takes a total of 6 hours, then we must solve

28/(16 - R) + 28/(5 + R) = 6

Solving this for R gives us

28*(5 + R) + 28*(16 - R) = 6*(16 - R)*(5 + R)
588 = 6*(80 + 11R - R^2)
98 = 80 + 11R - R^2
R^2 - 11R + 18 = 0
(R - 2)(R - 9) = 0
R = 2 or R = 9

Mathematically a river speed of either 2 mph or 9 mph can be the solution to the problem, although if you know something about rivers, you may know that 9 mph is incredibly fast for a river on which a small motorboat could run; 2 mph is more realistic.

(3) Dropping a Ball from a Building -- Easy

A bowling ball is dropped from the top of a building that is 31.6 meters tall. Because of the size of the ball and the low height of the building, we can use a simplified equation that ignores the effects of air resistance on a falling body. The height of the object (in meters) at time T is given by the equation

H(t) = 31.6 - 4.9T^2

The velocity (in meters per second) of the ball at any time T is given by the equation

V(T) = -9.8T

The negative sign is because the ball is falling down toward Earth.

How long does it take for the ball to hit the ground? How high is the ball when its velocity is -11 m/s? What is the average velocity of the ball from the time it's dropped to the time it lands?

Solution: To find when the ball hits the ground we solve the equation H(T) = 0. This gives us

31.6 - 4.9T^2 = 0
31.6 = 4.9T^2
31.6/4.9 = T^2
T^2 ≈ 6.44898
T ≈ 2.53948 seconds

To find out how high the ball is when its velocity is -11 m/s, we first solve V(T) = -11 for T, then plug that T value into the function H(T). This gives us

-9.8T = -11
T = 11/9.8
T ≈ 1.12245 seconds

= 31.6 - 4.9*(1.12245)^2
= 25.42652 meters

To find the average speed of the ball, we divide the height of the building by the number of seconds it takes for the ball to land. This gives us 31.6/2.53948 ≈ 12.44349.

Since the ball is going down, we need to add a negative sign. Therefore the average velocity is -12.44349 m/s.

(4) Price vs Demand and Maximum Profit -- Hard

Roy wants to produce and sell widgets. He knows that the demand for the widgets (i.e., the number of widgets he can sell) depends on the price. After doing some market research, he devises two possible functions that give demand D as a function of price P:

D1 = 1500 - 10P

D2 = 66000/(P + 20)

The total cost C to make N widgets is

C = 700 + 18N

which includes overhead and the cost of his time and materials.

At what price P do the two demand functions agree? Using the first demand function, how many widgets should he produce to maximize his total profits? For this second question, assume that he sells all the widgets he makes.

Solution: To find out what price P gives the same demand with either demand function, we set them equal to each other and solve for P. This gives us

1500 - 10P = 66000/(P + 20)
(1500 - 10P)(P + 20) = 66000
-10P^2 + 1300P + 30000 = 66000
10P^2 - 1300P + 36000 = 0
P^2 - 130P + 3600 = 0
(P - 40)(P - 90) = 0
P = 40 or P = 90

This means that at a price of $40 or $90 the two demand functions predict the same level of demand. At $40 the functions predict Roy can sell 1100 widgets. At a price of $90 the functions predict he can sell 600 widgets.

To find the optimal number of widgets to produce, we solve the first demand function for P, which gives us

P = 150 - D/10

Since demand is the number of widgets he can sell, let D = N. This let's us write

P = 150 - N/10.

Since profit equals total revenue minus total cost, and since total revenue is the price times the number sold, we have

Profit = (150 - N/10)*N - 700 - 18N
= -0.1N^2 + 132N - 700

The maximum value of this function is halfway between the roots. The roots are found by setting the quadratic equation equal to zero and solving for N.

-0.1N^2 + 132N - 700 = 0
N^2 - 1320N - 7000 = 0

Using the quadratic formula we get

N = 660 + 10*sqrt(4426)
N = 660 - 10*sqrt(4426)

The halfway point between these two roots is 660. Therefore, Roy should make 660 widgets and sell them for $84 each according to the first demand function. His total profits will be

84*660 - (700 + 18*660)
= $42,860


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    • FitnezzJim profile image

      FitnezzJim 18 months ago from Fredericksburg, Virginia

      It is interesting that the physics problems are the easy ones, and the business profit problems are the tougher ones. Is there a deeper lesson there?

    • Connie120 profile image

      Connie120 18 months ago

      That's very interesting. Quadratic equations were about the only things in algebra that I didn't hate! Everything else is too abstract.

    • profile image

      xandr 18 months ago

      How do you solve this one: One leg of a right triangle is 7 cm longer than the other. The perimeter is 40. Find the area of the triangle.

      So far i got was perimeter = 2x+7+h = 40 and area = x(x+7)/2. If i solve the perimeter equation for x i get x = (33-h)/2. But when i plug it into the area the hypotenuse variable doesn't cancel out.

    • calculus-geometry profile image

      TR Smith 18 months ago from Eastern Europe

      Hi Xandr, thanks for the question. Since you have a right triangle, h^2 =x^2 + (x+7)^2, or h = sqrt(2x^2 + 14x + 49). If you solve the perimeter equation, you get

      2x + 7 + h = 40

      sqrt(2x^2 + 14x + 49) = 33 - 2x

      2x^2 + 14x + 49 = 1089 - 132x + 4x^2

      0 = 2x^2 - 146x + 1040

      0 = x^2 - 73x + 520

      0 = (x - 65)(x - 8)

      x = 8 or x = 65

      You can see from the limits of the problem that x has to be 8. This means the legs of the triangle are 8 and 15, the hypotenuse is 17, and the area is 60

    • profile image

      Xandr 18 months ago

      Thank you

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