Applications of Quadratic Function

What are the dimensions of the largest rectangular field that can be enclosed by 80 m of fencing wire?

Solution:

Let land w be the length and width of a rectangle. Then, the perimeter P of a rectangle is

P = 2I+2. Since P = 80m thus,

2I + 2w = 80

I+w=40

I=40-w Expressing the length as a function of w

I=40-w Substituting in the formula for the area A of a rectangle

A(w) = wl

A(w) = w(40 - w)

A(w)=-w^2+40w

A(w) = - (w - 20) ^ 2 + 400

The vertex of the graph of the function A(w) is (20), 400). This point indicates a maximum value of 400 for A(w) that occurs when w = 20 Thus, the maximum area is 400 m² when the width is 20 m. If the width is 20 m, then the length is (40-20) m or 20 m also. The field with maximum area is a square.

Free falling objects can be modeled by a quadratic function h(t)=4.9+V+h, where h(t) is the height of an object at 1 seconds, when it is thrown with an initial velocity of V m/s and an initial height of h, meters. If the units are in feet, then the function is hir)=-16r+ Vt+h

From a 96-foot building, an object is thrown straight up into the air then follows a trajectory. The height S(t) of the ball above the building after 1 seconds is given by the function S(1)=801-16r.

Solution:

1. The maximum height reached by the object is the ordinate of the vertex of the parab ola of the function delta(t) = 80t - 16t ^ 2 By transforming this equation into the completed square form we have,

S(t) = 80t - 16t ^ 2

S(t) = - 16t ^ 2 + 80t

S(t) = - 16(t ^ 2 - 5t)

(t) = - 16(t ^ 2 - 5t + 25/4) + 100

S(t) = - 16 * (t - 5/2) ^ 2 + 100

The vertex is (5/2, 100) Thus the maximum height reached by the object is 100 ft from the top of the building. This is 196 ft from the ground.

The time for an object to reach the maximum height is the abscissa of the vertex of the parabola or the value of h.

S(t) = 80t - 16t ^ 2

S(t) = - 16(t - (5 ^ z)/2) + 100

Since the value of h is or 2.5, then the object is at its maximum height after 2.5 seconds.

To find the time it will take the object to hit the ground, let S(t) = - 96 , since the height of the building is 96 ft. The problem requires us to solve for 1.

h(t) = 80t - 16t ^ 2

- 96 = 80t - 16t ^ 2

16t ^ 2 - 80t - 96 = 0

t ^ 2 - 5t - 6 = 0

t-6) (t + 1) = 0

t = 6 t = - 1

Thus, it will take 6 seconds before the object hits the ground.

Problem: A garments store sells about 40 t-shirts per week at a price of Php 100 each. For each Php 10 decrease in price, the sales lady found out that 5 more t-shirts per week were sold. Write a quadratic function in standard form that models the revenue from t-shirt sales. What price produces the maximum revenue?

Solution:

Let x be the number of additional number of t-shirts sold.

You know that Revenue R(x) = (price per unit) x (number of units produced or sold). Therefore, Revenue R(x)=(Number of t-shirts sold) (Price per 1-shirt)

Revenue R(x) = (40 + 5x)(100 - 10x)

R(x) = - 50x ^ 2 + 100x + 4000

If we transform the function into the form y = a * (x - h) ^ 2 + k R(x) = - 50 * (x - 1) ^ 2 + 4050

The vertex is (1, 4050)

Thus, the maximum revenue is Php 4.050

The price of the t-shirt to produce maximum revenue can be determined by P(x) = 100 - 10x

P(x) = 100 - 10(1) = 90

Thus, Php 90 is the price of the t-shirt that produces maximum revenue.

Glossary of Terms

axis of symmetry- the vertical line through the vertex that divides the parabola into two equal parts direction of opening of a parabola can be determined from the value of a in f(x)= a * x ^ 2 + bx +c.lf a> 0, the parabola opens upward; if a < 0 the parabola opens downward.

domain of a quadratic function - the set of all possible values of x. Thus, the domain is the seto f all real numbers.

maximum value - the maximum value of f(x) = a * x ^ 2 + bx + c where a < 0 is the y-coordinate of the vertex.

minimum value - the minimum value of f(x) = a * x ^ 2 + bx + c where a > 0 , is the vertex.

Parabola the graph of a quadratic function.

quadratic function- a second-degree function of the form f(x) = a * x ^ 2 + bx + c where a, b, and e are real numbers and a ne0 This is a function which describes a polynomial of degree 2.

range of quadratic function-consists of all y greater than or equal to the y-coordinate of the vertex if the parabola opens upward.

- consists of all y less than or equal to the y-coordinate of the vertex if the parabola opens downward.

vertex - the turning point of the parabola or the lowest or highest point of the parabola. If the quadratic function is expressed in standard form y = a * (x - h) ^ 2 + k the vertex is the point (h, k)

zeros of a quadratic function-the values of x when y equals 0. The real zeros are the x-intercepts of the function's graph.