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



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mrbrianmclogan 4 years ago from Jacksonville

great job explaining! I really enjoy your hubs.

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