Solving Word Problems Involving Quadratic Characteristics Part Two Solving phrase Problems Involving Quadratic Functions Part Two

This hub is a sequel to the hub “Solving Word Problems involving Quadratic Functions. Within this hub I present three additional application difficulties. I hope you will enjoy this particular hub and gain much from it.

Sample Problem Number One : The sum of a pair of numbers is forty eight. Find the maximum merchandise and the two amounts.

Solution : Let n = the 1st number 48 – n = the other number f(n ) = product of these two numbers f(n ) = n (forty eight – n ) = 48n – n^2 f(n) = 48n - n ^2 Find the particular vertex of the parabola: a new = -1 (coefficient of n^2 b = forty eight (coefficient of n ) l = -b/2a =è this is the abscissa in the vertex of the parabola The parabola opens downward and the vertex is the maximum point since a can be a negative value. Find h : l = -48/2(-1) = 24 k = f(twenty-four ) = 48(twenty-four ) - 24^2 =1152 - 576 = 576 èthis could be the ordinate of the vertex. Thus the absolute maximum product is 576. The merchandise of the two numbers n and 48-n must be equal to the maximum merchandise 576. That is : n(forty eight – n) = 576 48n - n^2 = 576 n^2 -48n + 576 = 0 (n -24) ^2 = 0 n = twenty-four

If n = twenty-four then 48-n = 24. Therefore the a pair of numbers whose amount is 48 as well as whose product is a new maximum are twenty-four and 24.

Sample Problem Number Two : Find the dimensions and the maximum division of a rectangle if it is perimeter is thirty six. Solution ; Given : Let l = length Perimeter p = 36 P = 2l + 2w 36 = 2l + 2w w = ( thirty six -2l)/2 A = l* w A = ((36-2l )/2) l f(A) = 36l/2 - 2l^2/2 f(A) = 18l - l^2 a = -1 t = 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 a maximum therefore 81 is the maximum region. Solving pertaining to l : ( (36 – 2l)/2)l = 81 Multiplying this particular equation by two (36 – 2l) l = 162 36l – 2l^2 = 162 2l^2 -36l + 162 = zero Dividing by two l^2 -18l + 81 = 0 (l -9)^2 = 0 l=9 w = (thirty six - 2l)/2 = (36 -18)/2 = 18/2 = 9 The dimensions of the rectangle should be length = 9 and size = 9 so that you can get the maximum area.

Sample Problem Number Three : The sum of a pair of numbers is thirty six. Find the numbers whoever sum of the pieces is a minimum.

Solution: Let n = first number 36 –n = subsequent 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 picture by two F(S) = n^2 - 36n + 648 a new = 1 t = -36 h = -(-36)/2 = 18 k = farreneheit (18) = 18^2 – 36(16 ) + 648 k= 324 -648 + 648 = 324 This is the ordinate in the vertex The vertex of the parabola is a minimum since a new is a positive price. 324 could be the minimum sum of the particular squares 324 = n^2 – 36n + 648 n^2 -36n + 648 -324 = 0 n^2 -36n + 324 = 0 (n – 18)^2 = 0 n – 18 = zero ==è n = 18 The initial number is 16 and the other number is 18. Both numbers which will provide a minimum value for that sum of the pieces are 18 as well as 18.

SOURCE: ADVANCED ALGEBRA, TRIGONOMETRY as well as STATISTICS By : Orines Esparrago Reyes