Solving Word Problems Involving Quadratic Functions Part Two
Solving Word Problems Involving Quadratic Functions Part Two
This hub is a sequel to the hub “Solving Word Problems Involving Quadratic Functions. In this hub I present three additional application problems. I hope you will enjoy this hub and benefit much from it.
Sample Problem Number One :
The sum of two numbers is 48. Find the maximum product and the two numbers.
Solution :
Let n = The first number
48 – n = the second number
f(n) = product of the two numbers
f(n) = n (48 – n ) = 48n – n^2
f(n) = 48n - n ^2
Find the vertex of the parabola:
a = -1 (coefficient of n^2
b = 48 (coefficient of n)
h = -b/2a =è this is the abscissa of the vertex of the parabola
The parabola opens downward and the vertex is the maximum point since a is a negative value.
Find h : h = -48/2(-1) = 24
k = f(24) = 48(24) - 24^2 =1152 - 576 = 576 èthis is the ordinate of the vertex.
Thus the maximum product is 576. The product of the two numbers n and 48-n must be equal to the maximum product 576. That is :
n(48 – n) = 576
48n - n^2 = 576
n^2 -48n + 576 = 0
(n -24) ^2 = 0
n = 24
If n = 24 then 48-n = 24. Therefore the two numbers whose sum is 48 and whose product is a maximum are 24 and 24.
Sample Problem Number Two :
Find the dimension and the maximum area of a rectangle if its perimeter is 36.
Solution ;
Given :
Let l = length
Perimeter P = 36
P = 2l + 2w
36 = 2l + 2w
w = ( 36 -2l)/2
A = l* w
A = ((36-2l )/2) l
f(A) = 36l/2 - 2l^2/2
f(A) = 18l - l^2
a = -1 b = 18
h = -b/2a = -18/2(-1) = 9
k = f(9) = 18(9) - 9^2 = 162 - 81 = 81 è this is the ordinate of the parabola.
Vertex is at maximum therefore 81 is the maximum area.
Solving for l :
( (36 – 2l)/2)l = 81 Multiplying this equation by two
(36 – 2l) l = 162
36l – 2l^2 = 162
2l^2 -36l + 162 = 0 Dividing by two
l^2 -18l + 81 = 0
(l -9)^2 = 0
l = 9
w = (36 - 2l)/2 = (36 -18)/2 = 18/2 = 9
The dimensions of the rectangle should be length = 9 and width = 9 to be able to get the maximum area.
Sample Problem Number Three :
The sum of two numbers is 36. Find the numbers whose sum of the squares is a minimum.
Solution:
Let n = first number
36 –n = second number
F(S) = n^2 + (36 –n )^2
F(S) = n ^2 + 1,296 - 72n + n^2
F(S) = 2n^2 - 72n + 1,296 Dividing this equation by two
F(S) = n^2 - 36n + 648
a = 1 b = -36
h = -(-36)/2 = 18
k = f(18) = 18^2 – 36(18) + 648
k= 324 -648 + 648 = 324 This is the ordinate of the vertex
The vertex of the parabola is at minimum since a is a positive value.
324 is the minimum sum of the squares
324 = n^2 – 36n + 648
n^2 -36n + 648 -324 = 0
n^2 -36n + 324 = 0
(n – 18)^2 = 0
n – 18 = 0 ==è n = 18
The first number is 18 and the other number is 18. The two numbers which will give a minimum value for the sum of the squares are 18 and 18.
SOURCE:
ADVANCED ALGEBRA, TRIGONOMETRY AND STATISTICS
By :
Orines
Esparrago
Reyes
